  ### Home > CALC > Chapter 4 > Lesson 4.2.3 > Problem4-73s

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 = −9x from the area under y = 6x3. 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.