Let $a\left(x\right) = x^{3} − 3x$, $b\left(x\right) = a \left(x − 2\right) + 5$ and let $W = \int _ { 2 } ^ { 7 } a ( x ) d x$.
Recall $\int _ { 2 } ^ { 7 } a ( x ) d x$ means the area under the curve of $a\left(x\right)$ on $[2, 7]$.

1. Write $b\left(x\right)$ explicitly.

   Hint (a):

   Substitute $\left(x − 2\right)$ for every $x$ in a(x). Then add $5$.

2. Sketch the areas $\int _ { 2 } ^ { 7 } a ( x ) d x$ and $\int _ { 4 } ^ { 9 } b ( x ) d x$.

   Answer (b):

   

3. Find $\int _ { 4 } ^ { 9 } b ( x ) d x$ in terms of $W$.

   Answer (c):

   The Integral is moved $2$ to the right. But $b\left(x\right)$ is also moved $2$ to the right of $a\left(x\right)$. So that is a wash. But $b\left(x\right)$ is moved up $5$. So it has more area by its $\text{width} × 5$. In this case, the width is $5$. So if $W$ is the area under the '$a$' curve, then $W + 25$ is the area under the '$b$' curve.