Rewrite the following integral expressions as a single integral.  

1. $\int _ { - 5 } ^ { 2 } 6 x ^ { 3 } d x - \int _ { - 5 } ^ { 2 } - 9 x d x$

   Hint (a):

   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 $y = 6x^{3}$.

2. $\int _ { 5 } ^ { 9 } h ( x ) d x + \int _ { 9 } ^ { 3 } h ( x ) d x$

   Hint (b):

   Since both integrands are the same, $h\left(x\right)$, 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?

3. $\int _ { 2 } ^ { 7 } \pi ( 2 x ) ^ { 2 } d x + \int _ { 2 } ^ { 7 } \pi ( \sqrt { x } ) ^ { 2 } d x$

   Hint (c):

   Refer to the hint in part (a). Remember that you can factor out $π$.

   Answer (c):

   $\pi \int_{2}^{7}(2x)^{2}+(\sqrt{x})^{2}dx$

4. $\int _ { 1 } ^ { 4 } ( x + 5 ) ^ { 2 } d x + \int _ { 6 } ^ { 9 } \sqrt { x } d x$

   Hint (d):

   Notice that both the integrands and the bounds 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.

   Hint (d):

   You will not change the bounds, be sure to shift the function accordingly:
   $f\left(x − 5\right)$ will shift the function $5$ units to the right.
   $f\left(x + 5\right)$ will shift the function $5$ units to the left.