Evaluate the integral

Question

The graph of f(x) is given above on the interval [-1,11]. The areas bounded by the graph of f(x) and the x axis on the intervals [-1,2] , [2,6], and [ 6,10] are 4, 3, and 2 respectively. The function h(x) is defined on [-1,11] and given in the piecewise function above. The function g(x) is defined by:

g(x) = ∫ from 2 to x f(t) dt for -1 ≤ x ≤ 11

h(x) ={ x² + 3x -2 for x < -1

g(x), for -1 ≤ x ≤ 11

Evaluate the ∫ from 1 to 7 [3 h´(2x-4) +5] dx.

Answer 1

Hello Giselle! Because the only question explicitly stated in the problem is about solving the integral of h prime, I will ignore all information about f(x) and g(x).

∫(1, 7, 3 h´(2x - 4) + 5 , dx) = 3 * ∫(1, 7, h ' (2x - 4) , dx) + 5 ( 7 - 1 )

Notice that by u-substitution, you can set u = 2x - 4 and du / 2 = dx. Substituting these and changing the limits of integration to u (not to x) leads to..........

3 * ∫(1, 7, h ' (2x - 4) , dx) + 5 ( 7 - 1 ) = 1.5 * ∫( - 2, 10, h ' (u) , du ) + 30 =

1.5 * ( h(u) ) | (from -2 to 10) + 30

Antiderivative of h ' (u) is simply just h(u). Now, substitute u = -2 and u = 10 from the equation in the question for h ( - 2 ) and h( 10 ), which h( -2 ) = -4 and h (10) = g (10) (I don't know the equation of g(10), so you can figure that out), so......

1.5 * ( h(u) ) | (from -2 to 10) + 30 = 1.5 * ( g(10) + 4 ) + 30>>>>>>> (ANSWER) Hope this helped!