Math 19B Midterm Sol Extended.pdf - Solutions to midterm...

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Solutions to midterm, Math 19B, Spring 2016, with extended comments 1. (5 points) Compute the definite integral: Z - 1 - 3 ( 2 x - 5 xe x 2 + cos(3 x )) dx . Solution and comments: By linearity, we have Z - 1 - 3 3 x - 6 xe x 2 + cos(4 x ) dx = Z - 1 - 3 3 x dx - Z - 1 - 3 6 xe x 2 dx + Z - 1 - 3 cos(4 x ) dx. We will now evaluate each of the three integrals individually. 1. Recall that R 1 x dx = ln | x | + C . So, the definite integral is Z - 1 - 3 3 x dx = 3 Z - 1 - 3 1 x dx = 3 ln | x | - 1 - 3 = 3(ln | - 1 | - ln | - 3 | ) = 3(ln | 1 | - ln | 3 | ) = - 3 ln 3 = 3 ln 1 3 = ln 1 27 . If you had any of the the last five expressions then you are fine. 2. We will evaluate R 6 xe x 2 dx , via u -substitution. We let u = x 2 (I will say more about why u - substitution should probably be your first choice). Our differential is given as du = 2 x dx and so du 2 = x dx . Further, the bounds should be u ( - 3) = 9 and u ( - 1) = 1. Putting this all together we get: Z - 1 - 3 6 xe x 2 dx = 6 Z - 1 - 3 xe x 2 dx = 6 Z 1 9 e u du 2 = 3 Z 1 9 e u du = 3 e u 1 9 = 3( e - e 9 ) = 3 e (1 - e 8 ) . Either of the last two forms are fine. 3. There are two primary ways of doing this. The first is using the Fundamental Theorem of Calculus. The second is via u -substitution. For the first, note that the derivative of sin(4 x ) 4 is cos(4 x ). Thus, R cos(4 x ) dx = sin(4 x ) 4 + C . The second method is via u -substitution: let u = 4 x , then du = 4 dx and so du 4 = dx . Plugging everything in gives: R cos(4 x ) dx = R cos( u ) du 4 = sin( u ) 4 + C = sin(4 x ) 4 + C . So, the definite integral is Z - 1 - 3 cos(4 x ) dx = sin(4 x ) 4 - 1 - 3 ! = 1 4 (sin( - 4) - sin( - 12)) = 1 4 ( - sin(4) + sin(12)) . Either of the last two are fine. Also, you should note that neither of these last two are special values of the sine function. Combining all of these terms gives us Z - 1 - 3 3 x - 6 xe x 2 + cos(4 x ) dx = - 3 ln 3 - 3( e - e 9 ) + 1 4 (sin( - 4) - sin( - 12)) . Remark: 1
Use linearity. It is easier to evaluate several small integrals rather than one big integral. Or to put it another way, if you take one huge bite, you are more likely to choke than if you take several smaller bites. You need to know the rules about manipulating elementary functions. Specifically, the exponential, logarithm, trigonometric, and polynomials. You should know what some specific values are (e.g. e 0 = 1, cos( pi 2 ) = 0, etc.). You should know how to draw a rough graph of any of these functions. It does not have to be perfect, but it should be a qualitative graph that shows the major features. Be careful with your bounds with u -substitution: 3 R 1 9 e u du = 3( e u 1 9 ) is correct, but 3 R 1 9 e u du = 3 e u 1 9 = 3 e x 2 1 9 is not. Also, 3 R - 11 - 3 e u du is not correct. Be aware of what your bounds are. If you have a definite integral, then you should not have a + C at the end. The constant is there, because we can only distinguish antiderivatives up to a constant. While it may be difficult to distinguish when to use u -substitution and when to use integration by parts, there are some rules of thumb. If one of the functions, in the integrand, is a composition of two functions, then you may want to try u -substitution. If it is a product of two unrelated functions, then you may want to try integration by parts.

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