### Home > APCALC > Chapter 4 > Lesson 4.2.3 > Problem4-74

4-74.

If $\int _ { 2 } ^ { 4 } f ( x ) d x = 10$ evaluate:

1. $\int _ { 4 } ^ { 2 } - f ( x ) d x$

Notice that the bounds have flipped and the integrand is now negative.

2. $\int _ { 10 } ^ { 12 } ( f ( x - 8 ) + 4 ) d x$

Split this problem into a sum of two integrals:

$\int_{10}^{12}f(x-8)dx+\int_{10}^{12}4dx$

Integral 1: Notice that the bounds shifted $8$ units to the right... but so did the integrand: $f\left(x − 8\right)$.

$\int_{10}^{12}f(x-8)dx=\int_{2}^{4}f(x)dx=10.$

Integral 2: This integral represents a rectangular area with base: $12 − 10 = 2$ and height: $4$.

$\int_{10}^{12}4dx=(2)(4)=8$

3. $\int _ { 10 } ^ { 12 } f ( x + 8 ) d x$
The bounds shifted $8$ units to the right, but the integrand shifted $8$ units to the left.