# 022 Exam 2 Sample A, Problem 6

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

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

$(f\circ g)'(x)=f'(g(x))\cdot g'(x).$ The Product Rule: If $f$ and $g$ are differentiable functions, then

$(fg)'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x).$ The Quotient Rule: If $f$ and $g$ are differentiable functions and $g(x)\neq 0$ , then

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

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