sp08_152_Ch5_05

sp08_152_Ch5_05 - Substitution Rule Oguz Kurt OSU-Math 152...

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Unformatted text preview: Substitution Rule Oguz Kurt OSU-Math 152 Spring 2008 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 1/1 Area between y = f (x) and x-axis. Figure: y = f (x) Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 2/1 Area between y = f (x) and x-axis. Area = Figure: y = f (x) Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 2/1 Area between y = f (x) and x-axis. Area = a b |f (x)| dx Figure: y = f (x) Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 2/1 Area continued 1 Find the x-intercepts of f (x) on the interval [a, b]. Say, a c1 < c2 < .... < cr b are all intercepts. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 3/1 Area continued 1 2 Find the x-intercepts of f (x) on the interval [a, b]. Say, a c1 < c2 < .... < cr b are all intercepts. In each small interval, f (x) is either positive or negative. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 3/1 Area continued 1 2 Find the x-intercepts of f (x) on the interval [a, b]. Say, a c1 < c2 < .... < cr b are all intercepts. In each small interval, f (x) is either positive or negative. Hence, each small integral is either AREA or NEGATIVE AREA. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 3/1 Area continued 1 2 Find the x-intercepts of f (x) on the interval [a, b]. Say, a c1 < c2 < .... < cr b are all intercepts. In each small interval, f (x) is either positive or negative. Hence, each small integral is either AREA or NEGATIVE AREA. Absolute value function makes sure that negative values are made positive. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 3/1 Area continued 1 2 Find the x-intercepts of f (x) on the interval [a, b]. Say, a c1 < c2 < .... < cr b are all intercepts. In each small interval, f (x) is either positive or negative. Hence, each small integral is either AREA or NEGATIVE AREA. Absolute value function makes sure that negative values are made positive. Hence, you do no not have to worry about the sign of the absolute value before the integral. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 3/1 Area continued 1 2 Find the x-intercepts of f (x) on the interval [a, b]. Say, a c1 < c2 < .... < cr b are all intercepts. In each small interval, f (x) is either positive or negative. Hence, each small integral is either AREA or NEGATIVE AREA. Absolute value function makes sure that negative values are made positive. Hence, you do no not have to worry about the sign of the absolute value before the integral. Area = a b 3 |f (x)| dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 3/1 Area continued 1 2 Find the x-intercepts of f (x) on the interval [a, b]. Say, a c1 < c2 < .... < cr b are all intercepts. In each small interval, f (x) is either positive or negative. Hence, each small integral is either AREA or NEGATIVE AREA. Absolute value function makes sure that negative values are made positive. Hence, you do no not have to worry about the sign of the absolute value before the integral. Area = a b 3 |f (x)| dx |f (x)| dx + c2 c1 = a c1 |f (x)| dx + ... + b cr |f (x)| dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 3/1 Area continued 1 2 Find the x-intercepts of f (x) on the interval [a, b]. Say, a c1 < c2 < .... < cr b are all intercepts. In each small interval, f (x) is either positive or negative. Hence, each small integral is either AREA or NEGATIVE AREA. Absolute value function makes sure that negative values are made positive. Hence, you do no not have to worry about the sign of the absolute value before the integral. Area = a b 3 |f (x)| dx |f (x)| dx + f (x) dx + c2 c1 = a c1 |f (x)| dx + ... + c2 b cr |f (x)| dx b cr Spring 2008 3/1 = a Oguz Kurt (OSU-Math 152) c1 f (x) dx + ... + f (x) dx c1 Substitution Rule Area between f (x) and g(x) What happens if we would like to find the area between two functions? y=f(x) y=g(x) a 0 c d b Region between f(x) and g(x) on the closed interval [a,b] Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 4/1 Area between f (x) and g(x) on [a, b]: Area = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 5/1 Area between f (x) and g(x) on [a, b]: b a Area = |f (x) - g(x)| dx 1 Find the x-intercepts of f (x) - g(x) on the interval [a, b]. Say, a c1 < c2 < .... < cr b are all intercepts. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 5/1 Area between f (x) and g(x) on [a, b]: b a Area = |f (x) - g(x)| dx 1 Find the x-intercepts of f (x) - g(x) on the interval [a, b]. Say, a c1 < c2 < .... < cr b are all intercepts. Area = a b 2 |f (x) - g(x)| dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 5/1 Area between f (x) and g(x) on [a, b]: b a Area = |f (x) - g(x)| dx 1 Find the x-intercepts of f (x) - g(x) on the interval [a, b]. Say, a c1 < c2 < .... < cr b are all intercepts. Area = a b 2 |f (x) - g(x)| dx c2 c1 = a c1 |f (x)-g(x)| dx+ |f (x)-g(x)| dx+...+ b cr |f (x)-g(x)| dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 5/1 Area between f (x) and g(x) on [a, b]: b a Area = |f (x) - g(x)| dx 1 Find the x-intercepts of f (x) - g(x) on the interval [a, b]. Say, a c1 < c2 < .... < cr b are all intercepts. Area = a b 2 |f (x) - g(x)| dx c2 c1 = a c1 |f (x)-g(x)| dx+ [f (x)-g(x)]dx + |f (x)-g(x)| dx+...+ c2 b cr |f (x)-g(x)| dx b cr Spring 2008 5/1 = a c1 [f (x)-g(x)]dx +..+ [f (x)-g(x)]dx c1 Substitution Rule Oguz Kurt (OSU-Math 152) Example Find the area between the functions f (x) = x3 and g(x) = x on the closed interval [0,3]! Solution:. We will first find the x-intercepts of f (x) - g(x) = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 6/1 Example Find the area between the functions f (x) = x3 and g(x) = x on the closed interval [0,3]! Solution:. We will first find the x-intercepts of f (x) - g(x) = x3 - x = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 6/1 Example Find the area between the functions f (x) = x3 and g(x) = x on the closed interval [0,3]! Solution:. We will first find the x-intercepts of f (x) - g(x) = x3 - x = x(x - 1)(x + 1) = 0. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 6/1 Example Find the area between the functions f (x) = x3 and g(x) = x on the closed interval [0,3]! Solution:. We will first find the x-intercepts of f (x) - g(x) = x3 - x = x(x - 1)(x + 1) = 0. Hence, we have x = -1, 0, 1 as all of the x-intercepts. We will only use those in the interval [0, 3]. -1 is not in [0, 3] and 0 is already on the boundary. So, the only middle point that we have is x = 1. Now, the Area = 0 3 |x3 - x| dx = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 6/1 Example Find the area between the functions f (x) = x3 and g(x) = x on the closed interval [0,3]! Solution:. We will first find the x-intercepts of f (x) - g(x) = x3 - x = x(x - 1)(x + 1) = 0. Hence, we have x = -1, 0, 1 as all of the x-intercepts. We will only use those in the interval [0, 3]. -1 is not in [0, 3] and 0 is already on the boundary. So, the only middle point that we have is x = 1. Now, the Area = 0 1 0 3 |x3 - x| dx = (x3 - x) dx + Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 6/1 Example Find the area between the functions f (x) = x3 and g(x) = x on the closed interval [0,3]! Solution:. We will first find the x-intercepts of f (x) - g(x) = x3 - x = x(x - 1)(x + 1) = 0. Hence, we have x = -1, 0, 1 as all of the x-intercepts. We will only use those in the interval [0, 3]. -1 is not in [0, 3] and 0 is already on the boundary. So, the only middle point that we have is x = 1. Now, the Area = 0 1 0 3 |x3 - x| dx = 3 1 (x3 - x) dx + (x3 - x) dx = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 6/1 Example Find the area between the functions f (x) = x3 and g(x) = x on the closed interval [0,3]! Solution:. We will first find the x-intercepts of f (x) - g(x) = x3 - x = x(x - 1)(x + 1) = 0. Hence, we have x = -1, 0, 1 as all of the x-intercepts. We will only use those in the interval [0, 3]. -1 is not in [0, 3] and 0 is already on the boundary. So, the only middle point that we have is x = 1. Now, the Area = 0 1 0 3 |x3 - x| dx = 3 1 (x3 - x) dx + (x3 - x) dx = x4 x2 - 4 2 1 + 0 x4 x2 - 4 2 3 1 = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 6/1 Example Find the area between the functions f (x) = x3 and g(x) = x on the closed interval [0,3]! Solution:. We will first find the x-intercepts of f (x) - g(x) = x3 - x = x(x - 1)(x + 1) = 0. Hence, we have x = -1, 0, 1 as all of the x-intercepts. We will only use those in the interval [0, 3]. -1 is not in [0, 3] and 0 is already on the boundary. So, the only middle point that we have is x = 1. Now, the Area = 0 1 0 3 |x3 - x| dx = 3 1 (x3 - x) dx + (x3 - x) dx = x4 x2 - 4 2 1 + 0 x4 x2 - 4 2 3 1 = |(1/4 - 1/2) - 0| + Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 6/1 Example Find the area between the functions f (x) = x3 and g(x) = x on the closed interval [0,3]! Solution:. We will first find the x-intercepts of f (x) - g(x) = x3 - x = x(x - 1)(x + 1) = 0. Hence, we have x = -1, 0, 1 as all of the x-intercepts. We will only use those in the interval [0, 3]. -1 is not in [0, 3] and 0 is already on the boundary. So, the only middle point that we have is x = 1. Now, the Area = 0 1 3 |x3 - x| dx = 3 1 x4 x2 + - 4 2 0 0 1 = |(1/4 - 1/2) - 0| + |(81/4 - 9/2) - (1/4 - 1/2)| = (x3 - x) dx + (x3 - x) dx = Oguz Kurt (OSU-Math 152) Substitution Rule x4 x2 - 4 2 3 1 Spring 2008 6/1 Example Find the area between the functions f (x) = x3 and g(x) = x on the closed interval [0,3]! Solution:. We will first find the x-intercepts of f (x) - g(x) = x3 - x = x(x - 1)(x + 1) = 0. Hence, we have x = -1, 0, 1 as all of the x-intercepts. We will only use those in the interval [0, 3]. -1 is not in [0, 3] and 0 is already on the boundary. So, the only middle point that we have is x = 1. Now, the Area = 0 1 3 |x3 - x| dx = 3 1 x4 x2 + - 4 2 0 0 1 = |(1/4 - 1/2) - 0| + |(81/4 - 9/2) - (1/4 - 1/2)| = | - 1/4| + (x3 - x) dx + (x3 - x) dx = Oguz Kurt (OSU-Math 152) Substitution Rule x4 x2 - 4 2 3 1 Spring 2008 6/1 Example Find the area between the functions f (x) = x3 and g(x) = x on the closed interval [0,3]! Solution:. We will first find the x-intercepts of f (x) - g(x) = x3 - x = x(x - 1)(x + 1) = 0. Hence, we have x = -1, 0, 1 as all of the x-intercepts. We will only use those in the interval [0, 3]. -1 is not in [0, 3] and 0 is already on the boundary. So, the only middle point that we have is x = 1. Now, the Area = 0 1 3 |x3 - x| dx = 3 1 x4 x2 + - 4 2 0 0 1 = |(1/4 - 1/2) - 0| + |(81/4 - 9/2) - (1/4 - 1/2)| = | - 1/4| + |16| = (x3 - x) dx + (x3 - x) dx = Oguz Kurt (OSU-Math 152) Substitution Rule x4 x2 - 4 2 3 1 Spring 2008 6/1 Example Find the area between the functions f (x) = x3 and g(x) = x on the closed interval [0,3]! Solution:. We will first find the x-intercepts of f (x) - g(x) = x3 - x = x(x - 1)(x + 1) = 0. Hence, we have x = -1, 0, 1 as all of the x-intercepts. We will only use those in the interval [0, 3]. -1 is not in [0, 3] and 0 is already on the boundary. So, the only middle point that we have is x = 1. Now, the Area = 0 1 3 |x3 - x| dx = 3 1 x4 x2 + - 4 2 0 0 1 = |(1/4 - 1/2) - 0| + |(81/4 - 9/2) - (1/4 - 1/2)| = | - 1/4| + |16| = 16 + 1/4 = 16.25 (x3 - x) dx + (x3 - x) dx = Oguz Kurt (OSU-Math 152) Substitution Rule x4 x2 - 4 2 3 1 Spring 2008 6/1 Theorem: SUBSTITUTION RULE Let g(x) be differentiable and F (x) is an antiderivative of f (x). Then. if we set u = g(x) f (g(x))g (x) dx = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 7/1 Theorem: SUBSTITUTION RULE Let g(x) be differentiable and F (x) is an antiderivative of f (x). Then. if we set u = g(x) f (g(x))g (x) dx = f (u) du = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 7/1 Theorem: SUBSTITUTION RULE Let g(x) be differentiable and F (x) is an antiderivative of f (x). Then. if we set u = g(x) f (g(x))g (x) dx = f (u) du = F (u) + C = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 7/1 Theorem: SUBSTITUTION RULE Let g(x) be differentiable and F (x) is an antiderivative of f (x). Then. if we set u = g(x) f (g(x))g (x) dx = f (u) du = F (u) + C = F (g(x)) + C Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 7/1 Theorem: SUBSTITUTION RULE Let g(x) be differentiable and F (x) is an antiderivative of f (x). Then. if we set u = g(x) f (g(x))g (x) dx = f (u) du = F (u) + C = F (g(x)) + C Setting u = g(x) Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 7/1 Theorem: SUBSTITUTION RULE Let g(x) be differentiable and F (x) is an antiderivative of f (x). Then. if we set u = g(x) f (g(x))g (x) dx = f (u) du = F (u) + C = F (g(x)) + C Setting u = g(x) , we get du = g (x). dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 7/1 Theorem: SUBSTITUTION RULE Let g(x) be differentiable and F (x) is an antiderivative of f (x). Then. if we set u = g(x) f (g(x))g (x) dx = f (u) du = F (u) + C = F (g(x)) + C Setting u = g(x) , we get du = g (x) dx. du = g (x). This means that dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 7/1 Theorem: SUBSTITUTION RULE Let g(x) be differentiable and F (x) is an antiderivative of f (x). Then. if we set u = g(x) f (g(x))g (x) dx = f (u) du = F (u) + C = F (g(x)) + C du = g (x). This means that dx du = g (x) dx.Plugging in u = g(x) and du = g (x) dx, we get the second equation. Setting u = g(x) , we get Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 7/1 Theorem: SUBSTITUTION RULE Let g(x) be differentiable and F (x) is an antiderivative of f (x). Then. if we set u = g(x) f (g(x))g (x) dx = f (u) du = F (u) + C = F (g(x)) + C du = g (x). This means that dx du = g (x) dx.Plugging in u = g(x) and du = g (x) dx, we get the second equation. Setting u = g(x) , we get REMARK: We may see Substitution Rule as the reverse of Chain Rule. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 7/1 Example 1: (3x2 + 1) sin(x3 + x) dx If we set Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 8/1 Example 1: (3x2 + 1) sin(x3 + x) dx If we set u = x3 + x, Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 8/1 Example 1: (3x2 + 1) sin(x3 + x) dx If we set u = x3 + x, du/dx = (3x2 + 1). Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 8/1 Example 1: (3x2 + 1) sin(x3 + x) dx If we set u = x3 + x, du/dx = (3x2 + 1). So, du = (3x2 + 1)dx. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 8/1 Example 1: (3x2 + 1) sin(x3 + x) dx If we set u = x3 + x, du/dx = (3x2 + 1). So, du = (3x2 + 1)dx. (3x2 + 1) sin(x3 + x) dx = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 8/1 Example 1: (3x2 + 1) sin(x3 + x) dx If we set u = x3 + x, du/dx = (3x2 + 1). So, du = (3x2 + 1)dx. (3x2 + 1) sin(x3 + x) dx = = sin(x3 + x)(3x2 + 1) dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 8/1 Example 1: (3x2 + 1) sin(x3 + x) dx If we set u = x3 + x, du/dx = (3x2 + 1). So, du = (3x2 + 1)dx. (3x2 + 1) sin(x3 + x) dx = = = sin(x3 + x)(3x2 + 1) dx sin u du Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 8/1 Example 1: (3x2 + 1) sin(x3 + x) dx If we set u = x3 + x, du/dx = (3x2 + 1). So, du = (3x2 + 1)dx. (3x2 + 1) sin(x3 + x) dx = = = = sin(x3 + x)(3x2 + 1) dx sin u du - cos u + C Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 8/1 Example 1: (3x2 + 1) sin(x3 + x) dx If we set u = x3 + x, du/dx = (3x2 + 1). So, du = (3x2 + 1)dx. (3x2 + 1) sin(x3 + x) dx = = = = sin(x3 + x)(3x2 + 1) dx sin u du - cos u + C - cos(x3 + x) + C Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 8/1 Example 2: ln x dx x If we set Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 9/1 Example 2: ln x dx x If we set u = ln x, Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 9/1 Example 2: ln x dx x If we set u = ln x, du/dx = 1/x. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 9/1 Example 2: ln x dx x 1 dx. x If we set u = ln x, du/dx = 1/x. So, du = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 9/1 Example 2: ln x dx x 1 dx. x If we set u = ln x, du/dx = 1/x. So, du = ln x dx = x Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 9/1 Example 2: ln x dx x 1 dx. x 1 x dx If we set u = ln x, du/dx = 1/x. So, du = ln x dx = x = ln x Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 9/1 Example 2: ln x dx x 1 dx. x 1 x dx If we set u = ln x, du/dx = 1/x. So, du = ln x dx = x = = ln x u du Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 9/1 Example 2: ln x dx x 1 dx. x 1 x dx If we set u = ln x, du/dx = 1/x. So, du = ln x dx = x = = = ln x u du u2 +C 2 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 9/1 Example 2: ln x dx x 1 dx. x 1 x dx If we set u = ln x, du/dx = 1/x. So, du = ln x dx = x = = = ln x u du u2 +C 2 (ln x)2 +C 2 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 9/1 Example 2: If we set (4 tan + 5)4 sec2 d Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 10 / 1 Example 2: (4 tan + 5)4 sec2 d If we set u = 4 tan + 5, Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 10 / 1 Example 2: (4 tan + 5)4 sec2 d If we set u = 4 tan + 5, du/d = 4 sec2 . Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 10 / 1 Example 2: (4 tan + 5)4 sec2 d If we set u = 4 tan + 5, du/d = 4 sec2 . So, du = 4 sec2 d. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 10 / 1 Example 2: (4 tan + 5)4 sec2 d If we set u = 4 tan + 5, du/d = 4 sec2 . So, du = 4 sec2 d. (4 tan + 5)4 sec2 dx = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 10 / 1 Example 2: (4 tan + 5)4 sec2 d If we set u = 4 tan + 5, du/d = 4 sec2 . So, du = 4 sec2 d. (4 tan + 5)4 sec2 dx = (4 tan + 5)4 sec2 d Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 10 / 1 Example 2: (4 tan + 5)4 sec2 d If we set u = 4 tan + 5, du/d = 4 sec2 . So, du = 4 sec2 d. du (4 tan + 5)4 sec2 d = sec2 d (4 tan + 5)4 sec2 dx = 4 = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 10 / 1 Example 2: (4 tan + 5)4 sec2 d If we set u = 4 tan + 5, du/d = 4 sec2 . So, du = 4 sec2 d. du (4 tan + 5)4 sec2 d = sec2 d (4 tan + 5)4 sec2 dx = 4 du u4 = 4 = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 10 / 1 Example 2: (4 tan + 5)4 sec2 d If we set u = 4 tan + 5, du/d = 4 sec2 . So, du = 4 sec2 d. du (4 tan + 5)4 sec2 d = sec2 d (4 tan + 5)4 sec2 dx = 4 du u4 = 4 u5 +C = 20 = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 10 / 1 Example 2: (4 tan + 5)4 sec2 d If we set u = 4 tan + 5, du/d = 4 sec2 . So, du = 4 sec2 d. du (4 tan + 5)4 sec2 d = sec2 d (4 tan + 5)4 sec2 dx = 4 du u4 = 4 u5 +C = 20 (4 tan + 5)5 +C = 20 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 10 / 1 Example 2: (4 tan + 5)4 sec2 d If we set u = 4 tan + 5, du/d = 4 sec2 . So, du = 4 sec2 d. du (4 tan + 5)4 sec2 d = sec2 d (4 tan + 5)4 sec2 dx = 4 du u4 = 4 u5 +C = 20 (4 tan + 5)5 +C = 20 IMPORTANT REMARK: Most of the time, while doing substitution nothing about the previous variable should be left after substitution. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 10 / 1 Example 2: (4 tan + 5)4 sec2 d If we set u = 4 tan + 5, du/d = 4 sec2 . So, du = 4 sec2 d. du (4 tan + 5)4 sec2 d = sec2 d (4 tan + 5)4 sec2 dx = 4 du u4 = 4 u5 +C = 20 (4 tan + 5)5 +C = 20 IMPORTANT REMARK: Most of the time, while doing substitution nothing about the previous variable should be left after substitution. The only cases where this is not correct requires some algebraic manipulations. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 10 / 1 Example 2: (4 tan + 5)4 sec2 d If we set u = 4 tan + 5, du/d = 4 sec2 . So, du = 4 sec2 d. du (4 tan + 5)4 sec2 d = sec2 d (4 tan + 5)4 sec2 dx = 4 du u4 = 4 u5 +C = 20 (4 tan + 5)5 +C = 20 IMPORTANT REMARK: Most of the time, while doing substitution nothing about the previous variable should be left after substitution. The only cases where this is not correct requires some algebraic manipulations.Unless you found a way of manipulation, your substitution is incorrect either because of a mistake or because of the wrong choice. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 10 / 1 Example 3: If we set x x + 1 dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 11 / 1 Example 3: x x + 1 dx If we set u = x + 1, Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 11 / 1 Example 3: x x + 1 dx If we set u = x + 1, du/dx = 1. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 11 / 1 Example 3: x x + 1 dx If we set u = x + 1, du/dx = 1. So, du = dx. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 11 / 1 Example 3: x x + 1 dx If we set u = x + 1, du/dx = 1. So, du = dx. x x + 1 dx = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 11 / 1 Example 3: x x + 1 dx If we set u = x + 1, du/dx = 1. So, du = dx. x x + 1 dx = x x + 1 dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 11 / 1 Example 3: x x + 1 dx If we set u = x + 1, du/dx = 1. So, du = dx. x x + 1 dx = = x x + 1 dx x=u-1 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 11 / 1 Example 3: x x + 1 dx If we set u = x + 1, du/dx = 1. So, du = dx. x x + 1 dx = = = x x + 1 dx (u - 1) u du x=u-1 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 11 / 1 Example 3: x x + 1 dx If we set u = x + 1, du/dx = 1. So, du = dx. x x + 1 dx = = = = x x + 1 dx (u - 1) u du u3/2 - u1/2 du x=u-1 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 11 / 1 Example 3: x x + 1 dx If we set u = x + 1, du/dx = 1. So, du = dx. x x + 1 dx = = = = x x + 1 dx (u - 1) u du u3/2 - u1/2 du u5/2 - (5/2) x=u-1 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 11 / 1 Example 3: x x + 1 dx If we set u = x + 1, du/dx = 1. So, du = dx. x x + 1 dx = = = = = x x + 1 dx (u - 1) u du u3/2 - u1/2 du u5/2 u3/2 - +C (5/2) (3/2) x=u-1 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 11 / 1 Example 3: x x + 1 dx If we set u = x + 1, du/dx = 1. So, du = dx. x x + 1 dx = = = = = x x + 1 dx (u - 1) u du u3/2 - u1/2 du u5/2 u3/2 - +C (5/2) (3/2) 2(x + 1)5/2 - 5 x=u-1 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 11 / 1 Example 3: x x + 1 dx If we set u = x + 1, du/dx = 1. So, du = dx. x x + 1 dx = = = = = x x + 1 dx (u - 1) u du u3/2 - u1/2 du u5/2 u3/2 - +C (5/2) (3/2) 2(x + 1)5/2 2(x + 1)3/2 - +C 5 3 x=u-1 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 11 / 1 Substitution Rule for Definite Integrals Let g(x) be differentiable on [a, b] and f (x) be continuous on the range of g(x). Setting u = g(x) b a f (g(x))g (x) dx = g(b) g(a) f (u) du Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 12 / 1 Substitution Rule for Definite Integrals Let g(x) be differentiable on [a, b] and f (x) be continuous on the range of g(x). Setting u = g(x) b a f (g(x))g (x) dx = g(b) g(a) f (u) du If F (u) = f (u) du is an antiderivative of f (u) Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 12 / 1 Substitution Rule for Definite Integrals Let g(x) be differentiable on [a, b] and f (x) be continuous on the range of g(x). Setting u = g(x) b a f (g(x))g (x) dx = g(b) g(a) f (u) du If F (u) = b a f (u) du is an antiderivative of f (u) then f (g(x))g (x) dx = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 12 / 1 Substitution Rule for Definite Integrals Let g(x) be differentiable on [a, b] and f (x) be continuous on the range of g(x). Setting u = g(x) b a f (g(x))g (x) dx = g(b) g(a) f (u) du If F (u) = b a f (u) du is an antiderivative of f (u) then f (g(x))g (x) dx = [F (g(x)]b = a Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 12 / 1 Substitution Rule for Definite Integrals Let g(x) be differentiable on [a, b] and f (x) be continuous on the range of g(x). Setting u = g(x) b a f (g(x))g (x) dx = g(b) g(a) f (u) du If F (u) = b a f (u) du is an antiderivative of f (u) then g(b) g(b) g(a) f (g(x))g (x) dx = [F (g(x)]b = [F (u)]g(a) = a f (u) du Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 12 / 1 Substitution Rule for Definite Integrals Let g(x) be differentiable on [a, b] and f (x) be continuous on the range of g(x). Setting u = g(x) b a f (g(x))g (x) dx = g(b) g(a) f (u) du If F (u) = b a f (u) du is an antiderivative of f (u) then g(b) g(b) g(a) f (g(x))g (x) dx = [F (g(x)]b = [F (u)]g(a) = a g(b) g(a) f (u) du : REMARK: You do not have to use f (u) du. One can first find the indefinite integral and then plug-in as well. Surely, use of this rule saves some space and time. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 12 / 1 Method 1: If we set 5 0 x 9 - x2 dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 13 / 1 Method 1: 5 0 x 9 - x2 dx If we set u = 9 - x2 , Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 13 / 1 Method 1: 5 0 x 9 - x2 dx If we set u = 9 - x2 , du/dx = -2x. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 13 / 1 Method 1: 5 0 x 9 - x2 dx If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 13 / 1 Method 1: 5 0 x 9 - x2 dx If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. 5 0 x 9 - x2 dx = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 13 / 1 Method 1: 5 0 x 9 - x2 dx 5 If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. 5 0 x 9 - x2 dx = 0 ( 9 - x2 )x dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 13 / 1 Method 1: 5 0 x 9 - x2 dx 5 If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. 5 0 x 9 - x2 dx = = 0 ( 9 - x2 )x dx x dx = du -2 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 13 / 1 Method 1: 5 0 x 9 - x2 dx 5 If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. 5 0 x 9 - x2 dx = = 0 ( 9 - x2 )x dx du -2 x dx = du -2 u Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 13 / 1 Method 1: 5 0 x 9 - x2 dx 5 If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. 5 0 x 9 - x2 dx = = = 0 ( 9 - x2 )x dx du -2 x dx = du -2 u (What about the end points?) Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 13 / 1 Method 1: 5 0 x 9 - x2 dx 5 If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. 5 0 x 9 - x2 dx = = = = 0 ( 9 - x2 )x dx g( 5) x dx = 1 - 2 u du -2 du -2 (What about the end points?) u1/2 du g(0) = 9, g( 5) = 4 g(0) Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 13 / 1 Method 1: 5 0 x 9 - x2 dx 5 If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. 5 0 x 9 - x2 dx = = = = 0 ( 9 - x2 )x dx g( 5) x dx = 1 - 2 - 1 2 u du -2 du -2 (What about the end points?) u1/2 du g(0) = 9, g( 5) = 4 g(0) 4 9 u1/2 du = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 13 / 1 Method 1: 5 0 x 9 - x2 dx 5 If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. 5 0 x 9 - x2 dx = = = = = 0 ( 9 - x2 )x dx g( 5) x dx = 1 - 2 - 1 2 u du -2 du -2 (What about the end points?) u1/2 du 1 2 g(0) = 9, g( 5) = 4 9 g(0) 4 9 u1/2 du = - - u1/2 du 4 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 13 / 1 Method 1: 5 0 x 9 - x2 dx 5 If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. 5 0 x 9 - x2 dx = = = = = 0 ( 9 - x2 )x dx g( 5) x dx = 1 - 2 - 1 2 u du -2 du -2 (What about the end points?) u1/2 du 1 2 g(0) = 9, g( 5) = 4 9 g(0) 4 9 u1/2 du = - 9 - u1/2 du 4 1 2 (3/2) u3/2 = 4 Spring 2008 13 / 1 Oguz Kurt (OSU-Math 152) Substitution Rule Method 1: 5 0 x 9 - x2 dx 5 If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. 5 0 x 9 - x2 dx = = = = = 0 ( 9 - x2 )x dx g( 5) x dx = 1 - 2 - 1 2 u du -2 du -2 (What about the end points?) u1/2 du 1 2 g(0) = 9, g( 5) = 4 9 g(0) 4 9 u1/2 du = - 9 - u1/2 du 4 1 2 (3/2) u3/2 = 4 93/2 3 Spring 2008 13 / 1 Oguz Kurt (OSU-Math 152) Substitution Rule Method 1: 5 0 x 9 - x2 dx 5 If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. 5 0 x 9 - x2 dx = = = = = 0 ( 9 - x2 )x dx g( 5) x dx = 1 - 2 - 1 2 u du -2 du -2 (What about the end points?) u1/2 du 1 2 g(0) = 9, g( 5) = 4 9 g(0) 4 9 u1/2 du = - 9 - 3 = u1/2 du 4 1 2 (3/2) u3/2 = 4 93/2 3 - 43/2 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 13 / 1 Method 1: 5 0 x 9 - x2 dx 5 If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. 5 0 x 9 - x2 dx = = = = = 0 ( 9 - x2 )x dx g( 5) x dx = 1 - 2 - 1 2 u du -2 du -2 (What about the end points?) u1/2 du 1 2 g(0) = 9, g( 5) = 4 9 g(0) 4 9 u1/2 du = - 9 - 3 = u1/2 du 4 1 2 (3/2) u3/2 = 4 93/2 3 - 43/2 27 3 Spring 2008 13 / 1 Oguz Kurt (OSU-Math 152) Substitution Rule Method 1: 5 0 x 9 - x2 dx 5 If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. 5 0 x 9 - x2 dx = = = = = 0 ( 9 - x2 )x dx g( 5) x dx = 1 - 2 - 1 2 u du -2 du -2 (What about the end points?) u1/2 du 1 2 g(0) = 9, g( 5) = 4 9 g(0) 4 9 u1/2 du = - 9 - 3 = u1/2 du 4 1 2 (3/2) u3/2 = 4 93/2 3 - 43/2 27 8 - = 3 3 Spring 2008 13 / 1 Oguz Kurt (OSU-Math 152) Substitution Rule Method 1: 5 0 x 9 - x2 dx 5 If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. 5 0 x 9 - x2 dx = = = = = 0 ( 9 - x2 )x dx g( 5) x dx = 1 - 2 - 1 2 u du -2 du -2 (What about the end points?) u1/2 du 1 2 g(0) = 9, g( 5) = 4 9 g(0) 4 9 u1/2 du = - 9 - 3 = u1/2 du 4 1 2 (3/2) u3/2 = 4 93/2 3 - 43/2 27 8 19 - = 3 3 3 Spring 2008 13 / 1 Oguz Kurt (OSU-Math 152) Substitution Rule Method 2: 5 0 x 9 - x2 dx First find the indefinite integral. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 14 / 1 Method 2: 5 0 x 9 - x2 dx First find the indefinite integral. If we set Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 14 / 1 Method 2: 5 0 x 9 - x2 dx First find the indefinite integral. If we set u = 9 - x2 , Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 14 / 1 Method 2: 5 0 x 9 - x2 dx First find the indefinite integral. If we set u = 9 - x2 , du/dx = -2x. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 14 / 1 Method 2: 5 0 x 9 - x2 dx First find the indefinite integral. If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 14 / 1 Method 2: 5 0 x 9 - x2 dx First find the indefinite integral. If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. x 9 - x2 dx = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 14 / 1 Method 2: 5 0 x 9 - x2 dx First find the indefinite integral. If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. x 9 - x2 dx = ( 9 - x2 )x dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 14 / 1 Method 2: 5 0 x 9 - x2 dx First find the indefinite integral. If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. x 9 - x2 dx = = ( 9 - x2 )x dx x dx = du -2 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 14 / 1 Method 2: 5 0 x 9 - x2 dx First find the indefinite integral. If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. x 9 - x2 dx = = ( 9 - x2 )x dx u du = -2 x dx = du -2 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 14 / 1 Method 2: 5 0 x 9 - x2 dx First find the indefinite integral. If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. x 9 - x2 dx = = = ( 9 - x2 )x dx u x dx = du -2 du 1 = - -2 2 u1/2 du Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 14 / 1 Method 2: 5 0 x 9 - x2 dx First find the indefinite integral. If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. x 9 - x2 dx = = = ( 9 - x2 )x dx u x dx = du -2 du 1 = - -2 2 3/2 1 u - +C = 2 (3/2) u1/2 du Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 14 / 1 Method 2: 5 0 x 9 - x2 dx First find the indefinite integral. If we set u = 9 - x2 , du/dx = -2x. So, du = -2x dx. x 9 - x2 dx = = = ( 9 - x2 )x dx u x dx = du -2 du 1 = - u1/2 du -2 2 1 u3/2 -(9 - x2 )3/2 - +C = +C 2 (3/2) 3 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 14 / 1 Method 2 continued Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 15 / 1 Method 2 continued 0 5 x 9 - x2 dx = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 15 / 1 Method 2 continued 0 5 x 9- x2 dx = = -(9 - x2 )3/2 3 5 0 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 15 / 1 Method 2 continued 0 5 x 9- x2 dx = = = -(9 - x2 )3/2 3 -(9 - 3 5)3/2 5 0 - -(9 - 0)3/2 3 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 15 / 1 Method 2 continued 0 5 x 9- x2 dx = = = -(9 - x2 )3/2 3 -(9 - 3 - 43/2 3 5)3/2 5 0 - -(9 - 0)3/2 3 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 15 / 1 Method 2 continued 0 5 x 9- x2 dx = = = ... ... = -(9 - x2 )3/2 3 -(9 - 3 - 43/2 3 5)3/2 5 0 - 93/2 3 -(9 - 0)3/2 3 + = ... Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 15 / 1 Method 2 continued 0 5 x 9- x2 dx = = = ... ... = -(9 - x2 )3/2 3 -(9 - 3 - 43/2 3 5)3/2 5 0 - 93/2 3 -(9 - 0)3/2 3 + = ... 19 3 Substitution Rule Spring 2008 15 / 1 Oguz Kurt (OSU-Math 152) IMPORTANT REMARK Some forget to plug in u = g(x) back to its original place before they plug in the end points and make the following mistake: Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 16 / 1 IMPORTANT REMARK Some forget to plug in u = g(x) back to its original place before they plug in the end points and make the following mistake: -u3/2 3 0 5 = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 16 / 1 IMPORTANT REMARK Some forget to plug in u = g(x) back to its original place before they plug in the end points and make the following mistake: -u3/2 3 0 5 -( 5)3/2 = -0= 3 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 16 / 1 IMPORTANT REMARK Some forget to plug in u = g(x) back to its original place before they plug in the end points and make the following mistake: -u3/2 3 0 5 -( 5)3/2 -53/4 = -0= 3 3 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 16 / 1 IMPORTANT REMARK Some forget to plug in u = g(x) back to its original place before they plug in the end points and make the following mistake: -u3/2 3 0 5 -( 5)3/2 -53/4 19 = -0= = 3 3 3 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 16 / 1 IMPORTANT REMARK Some forget to plug in u = g(x) back to its original place before they plug in the end points and make the following mistake: -u3/2 3 0 5 -( 5)3/2 -53/4 19 = -0= = 3 3 3 BAD!!!!!! Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 16 / 1 IMPORTANT REMARK Some forget to plug in u = g(x) back to its original place before they plug in the end points and make the following mistake: -u3/2 3 0 5 -( 5)3/2 -53/4 19 = -0= = 3 3 3 BAD!!!!!! though, in reality, we cannot plug in a value of the variable x in place of a variable u. WHY? Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 16 / 1 IMPORTANT REMARK Some forget to plug in u = g(x) back to its original place before they plug in the end points and make the following mistake: -u3/2 3 0 5 -( 5)3/2 -53/4 19 = -0= = 3 3 3 BAD!!!!!! though, in reality, we cannot plug in a value of the variable x in place of a variable u. WHY? Because we have Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 16 / 1 IMPORTANT REMARK Some forget to plug in u = g(x) back to its original place before they plug in the end points and make the following mistake: -u3/2 3 0 5 -( 5)3/2 -53/4 19 = -0= = 3 3 3 BAD!!!!!! though, in reality, we cannot plug in a value of the variable x in place of a variable u. WHY? Because we have -u3/2 x= 5 3 x=0 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 16 / 1 SYMMETRY THEOREM Definition: f (x) is called EVEN if f (-x) = f (x) Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 17 / 1 SYMMETRY THEOREM Definition: f (x) is called EVEN if f (-x) = f (x) ODD if f (-x) = -f (x) Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 17 / 1 SYMMETRY THEOREM Definition: f (x) is called EVEN if f (-x) = f (x) ODD if f (-x) = -f (x) SYMMETRY THEOREM: Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 17 / 1 SYMMETRY THEOREM Definition: f (x) is called EVEN if f (-x) = f (x) ODD if f (-x) = -f (x) SYMMETRY THEOREM: If f (x) is EVEN Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 17 / 1 SYMMETRY THEOREM Definition: f (x) is called EVEN if f (-x) = f (x) ODD if f (-x) = -f (x) SYMMETRY THEOREM: If f (x) is EVEN then a -a f (x) dx = 2 0 a f (x) dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 17 / 1 SYMMETRY THEOREM Definition: f (x) is called EVEN if f (-x) = f (x) ODD if f (-x) = -f (x) SYMMETRY THEOREM: If f (x) is EVEN then If f (x) is ODD a -a f (x) dx = 2 0 a f (x) dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 17 / 1 SYMMETRY THEOREM Definition: f (x) is called EVEN if f (-x) = f (x) ODD if f (-x) = -f (x) SYMMETRY THEOREM: If f (x) is EVEN then If f (x) is ODD then a -a a -a f (x) dx = 2 0 a f (x) dx f (x) dx = 0 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 17 / 1 SYMMETRY THEOREM Definition: f (x) is called EVEN if f (-x) = f (x) ODD if f (-x) = -f (x) SYMMETRY THEOREM: If f (x) is EVEN then If f (x) is ODD then a -a a -a f (x) dx = 2 0 a f (x) dx f (x) dx = 0 Note that this theorem can NOT be used if f (x) is neither odd nor even. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 17 / 1 WHY? Setting u = -x Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 18 / 1 WHY? Setting u = -x(which implies du = -dx) Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 18 / 1 WHY? Setting u = -x(which implies du = -dx), we get 0 -a 0 a f (x) dx = - f (-u) du = + f (-u) du a 0 f (x) is even [f (-u) = f (u)] 0 -a f (x) dx = 0 a f (u) du f (x) is odd [f (-u) = -f (u)] 0 -a f (x)dx = - a 0 f (u)du Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 18 / 1 SYMMETRY THEOREM CONT'D if f is EVEN then Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 19 / 1 SYMMETRY THEOREM CONT'D if f is EVEN then a -a f (x) dx = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 19 / 1 SYMMETRY THEOREM CONT'D if f is EVEN then a -a 0 -a a 0 f (x) dx = = f (x) dx + f (x) dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 19 / 1 SYMMETRY THEOREM CONT'D if f is EVEN then a -a 0 -a a 0 a 0 f (x) dx = = f (x) dx + f (x) dx f (x) dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 19 / 1 SYMMETRY THEOREM CONT'D if f is EVEN then a -a 0 -a a 0 a 0 a 0 f (x) dx = = = f (x) dx + f (x) dx + f (x) dx f (x) dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 19 / 1 SYMMETRY THEOREM CONT'D if f is EVEN then a -a 0 -a a 0 a 0 a 0 f (x) dx = = = 2 f (x) dx + f (x) dx + a f (x) dx f (x) dx f (x) dx 0 if f is ODD then Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 19 / 1 SYMMETRY THEOREM CONT'D if f is EVEN then a -a 0 -a a 0 a 0 a 0 f (x) dx = = = 2 f (x) dx + f (x) dx + a f (x) dx f (x) dx f (x) dx 0 if f is ODD then a -a f (x) dx = Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 19 / 1 SYMMETRY THEOREM CONT'D if f is EVEN then a -a 0 -a a 0 a 0 a 0 f (x) dx = = = 2 f (x) dx + f (x) dx + a f (x) dx f (x) dx f (x) dx 0 if f is ODD then a -a f (x) dx = = 0 -a f (x) dx + 0 a f (x) dx Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 19 / 1 SYMMETRY THEOREM CONT'D if f is EVEN then a -a 0 -a a 0 a 0 a 0 f (x) dx = = = 2 f (x) dx + f (x) dx + a f (x) dx f (x) dx f (x) dx 0 if f is ODD then a -a f (x) dx = = - 0 -a f (x) dx + 0 a a f (x) dx f (x) dx 0 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 19 / 1 SYMMETRY THEOREM CONT'D if f is EVEN then a -a 0 -a a 0 a 0 a 0 f (x) dx = = = 2 f (x) dx + f (x) dx + a f (x) dx f (x) dx f (x) dx 0 if f is ODD then a -a f (x) dx = = - = 0 -a f (x) dx + 0 a a f (x) dx a f (x) dx + 0 f (x) dx 0 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 19 / 1 SYMMETRY THEOREM CONT'D if f is EVEN then a -a 0 -a a 0 a 0 a 0 f (x) dx = = = 2 f (x) dx + f (x) dx + a f (x) dx f (x) dx f (x) dx 0 if f is ODD then a -a f (x) dx = = - = 0 0 -a f (x) dx + 0 a a f (x) dx a f (x) dx + 0 f (x) dx 0 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 19 / 1 Examples: /2 1 sin x dx = -/2 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 20 / 1 Examples: /2 1 sin x dx = 0 -/2 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 20 / 1 Examples: /2 1 -/2 sin x dx = 0 because sin(-x) = - sin x (ODD) Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 20 / 1 Examples: /2 1 -/2 /2 2 sin x dx = 0 because sin(-x) = - sin x (ODD) cos x dx = -/2 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 20 / 1 Examples: /2 1 -/2 /2 2 sin x dx = 0 because sin(-x) = - sin x (ODD) cos x dx = 2 0 /2 cos x dx -/2 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 20 / 1 Examples: /2 1 -/2 /2 2 sin x dx = 0 because sin(-x) = - sin x (ODD) cos x dx = 2 0 /2 cos x dx because cos(-x) = cos x (EVEN) -/2 Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 20 / 1 Examples: /2 1 -/2 /2 2 sin x dx = 0 because sin(-x) = - sin x (ODD) cos x dx = 2 0 1 -1 /2 cos x dx because cos(-x) = cos x (EVEN) -/2 3 EXC: (v cos v + sin3 v + v 3 )dv =? Hint: Check whether the function is odd or even. Oguz Kurt (OSU-Math 152) Substitution Rule Spring 2008 20 / 1 ...
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