MATH
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# The antiderivatives of g 1 x g 2 x only differ by a

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The antiderivatives of G 1 (x) - G 2 (x) only differ by a constant C. This means that our starting values of 1 and 3 can be plugged into the places of each antiderivatives. This becomes: 0 5 3 1 dx which is 0 5 2 dx . This gives us the answer 10.

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3. In this problem we practice sigma notation and the definition of integrals. For example, consider the integral 1 3 x 2 dx . Let's use sigma notation to write an approximation of this integral given by a right-hand sum with 5 rectangles. The width of each rectangle is x = 3 1 5 = 2 5 = .4 . From the picture below we see that the right endpoint of the k th rectangle is x k = 1 + 2 k 5 = 1 + .4 k . Putting this together we have Right(5) = k = 1 5 x k 2 x = k = 1 5 ( 1 + 2 k 5 ) 2 5 . If we wanted to find the sum with n subdivisions, we get a an expression for Right( n ), Right( n ) = k = 1 n ( 1 + 2 k n ) 2 n . Note that the more rectangles we use, the more accurate our approximation of the integral. In fact, by the definition of a definite integral, 1 + 2 k n ¿ 2 2 n ¿ lim n→∞ k = 1 n ¿ = 1 3 x 2 dx.
Use this example as a model when answering the following questions. a. Write an expression using sigma notation for a right-hand approximation with n subdivisions to the following integrals. i. 0 1 x 2 dx = k = 0 n x 2 1 n In this equation, Δ x = 1 0 n = 1 n and the value of k is equal to the left endpoint on the interval (0, 1). ii. 2 6 x 3 dx = k = 2 n x 3 4 n In this equation, Δ x = 6 2 n = 4 n and the value of k is equal to the left endpoint on the interval (2, 6) b. Compute the following limit: lim n→∞ 1 n 3 k = 1 n k 2 To find the limit of this equation we have to evaluate the summation. To do this, you first have to recognize that the summation of k squared is equivalent to n ( n + 1 )( 4 n + 1 ) 6 using the definition of a definite integral. Next, you simplify the expression and distribute the the 1 n 3 . The final answer is 4 n 2 + 4 n + 1 6 n 2 . Hint: This is related to your answer to part (a). c. Find a and b which make the following identity hold: lim n→ ∞ 1 n 3 k = 1 n 10 n e 1 + 10 k / n = a b e x dx . For the given identity, the value of a is 1, and the value of b is 11. The value of a is equivalent to the k value because that is where you start your subintervals. The value of b was found using the

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equation Δ x = b a n where a =1 and Δ x = 10 n . When you plug in those values you can solve for b. 4. Let f ( x ) be the function graphed below. Note that for 6 ≤x ≤ 8 , f ( x ) is in the shape of a semicircle. a. Carefully sketch the graph of F 1 ( x ) , a continuous antiderivative of f ( x ) passing through ( 3,0 ) . Clearly label all local extrema and inflection points in your graph. The graph of F 1 ( x ) was found by using the first fundamental theorem of calculus to calculate each point individually. The concavity of the function was determined by the graph of the derivative. Everywhere that the derivation function was increasing, the graph of F 1 ( x ) is concave up and everywhere the the derivation function is decreasing F 1 ( x ) is concave down.
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