### Home > APCALC > Chapter 4 > Lesson 4.3.2 > Problem4-126

4-126.

Let $\int _ { 3 } ^ { 5 } h ( x ) d x = 4$ and $\int _ { 3 } ^ { 5 } j ( x ) d x = 2$. Evaluate:

1. $\int _ { 5 } ^ { 3 } ( h ( x ) + j ( x ) ) d x$

Notice that the bounds are reverse. What does that 'do' to the way we evaluate area?

2. $\int _ { 5 } ^ { 7 } ( h ( x - 2 ) + 2 ) d x$

Two shifts happen to the integrand: It is shifted two units up. It is shifted two units right. The bounds also changed.

Shifting up $2$ units is like adding the integral:

$\int_{5}^{7}2dx=4$

In other words, it's like adding a rectangle with height $2$ and base $\left(7 − 5\right)$.

Shifting an integral to the right could make it impossible to evaluate.
Fortunately, in this case, the bounds also shifted $2$ units to the right.

$\text{So }\int_{3}^{5}h(x)dx=\int_{5}^{7}h(x-2)dx=4$

$\text{Therefore: }\int_{5}^{7}h(x-2)+2=4+4=8$