### Home > CALC > Chapter 4 > Lesson 4.3.2 > Problem4-115

4-115.

Let $\int _ { 3 } ^ { 5 } h ( x ) d x = 4$ and $\int _ { 3 } ^ { 5 } j ( x ) d x = 2$. Find $\int _ { 5 } ^ { 3 } ( h ( x ) + j ( x ) ) d x$ and $\int _ { 5 } ^ { 7 } ( h ( x - 2 ) + 2 ) d x$.

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

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:
In other words, it's like adding a rectangle with height $2$ and base $(7-5)$.

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.