# 022 Exam 2 Sample A, Problem 6

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Find the area under the curve of  ${\displaystyle y\,=\,{\frac {8}{\sqrt {x}}}}$  between ${\displaystyle x=1}$ and ${\displaystyle x=4}$.

Foundations:
This problem requires two rules of integration. In particular, you need
Integration by substitution (U - sub): If ${\displaystyle f}$ and ${\displaystyle g}$ are differentiable functions, then

${\displaystyle (f\circ g)'(x)=f'(g(x))\cdot g'(x).}$

The Product Rule: If ${\displaystyle f}$ and ${\displaystyle g}$ are differentiable functions, then

${\displaystyle (fg)'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x).}$

The Quotient Rule: If ${\displaystyle f}$ and ${\displaystyle g}$ are differentiable functions and ${\displaystyle g(x)\neq 0}$ , then

${\displaystyle \left({\frac {f}{g}}\right)'(x)={\frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^{2}}}.}$
Additionally, we will need our power rule for differentiation:
${\displaystyle \left(x^{n}\right)'\,=\,nx^{n-1},}$ for ${\displaystyle n\neq 0}$,
as well as the derivative of natural log:
${\displaystyle \left(\ln x\right)'\,=\,{\frac {1}{x}}.}$

Solution:

Step 1:
Set up the integral:
${\displaystyle \int _{1}^{4}{\frac {8}{\sqrt {x}}}dx}$
Step 2:
Using the power rule we have:
$array}"): {\displaystyle \begin{array}{rcl} \int_1^4 \frac{8}{\sqrt{x}}dx & = & \frac{x^{1/2}} \end{array}$
Step 3:
Now we need to substitute back into our original variables using our original substitution ${\displaystyle u=3x+2}$
to get ${\displaystyle {\frac {\log(u)}{3}}={\frac {\log(3x+2}{3}}}$
Step 4:
Since this integral is an indefinite integral we have to remember to add "+ C" at the end.
${\displaystyle \int {\frac {1}{3x+2}}dx={\frac {\ln(3x+2)}{3}}+C}$