### Home > CALC > Chapter 4 > Lesson 4.2.3 > Problem 4-73s

Notice that the bounds are the same but the integrands are different. What this means is that, on the interval *x* = −5 through *x* = 2, we are subtracting the area under *y* = −9*x* from the area under *y* = 6*x*^{3}.

Since both integrands are the same, *h*(*x*), notice the bounds. The first integral has bounds that move forward, starting at *x* = 5 and

ending at *x* = 9. While the second integral has bounds that move backwards, from *x* = 9 to *x* = 3. What area remains?

Refer to the hint in part (a). Remember that you can factor out *π*.

Notice that both the integrands and the bouds are different. The only thing that these integrals have in common is the distance between the bounds: 4 − 1 = 9 − 6. We can shift the first integral 5 units to the right to match the second integral, or vice versa.

You will not change the bounds, be sure to shift the function accordingly:*f*(*x* − 5) will shift the function 5 units to the right.*f*(*x* + 5) will shift the function 5 units to the left.