# Difference between revisions of "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:
${\displaystyle {\begin{array}{rcl}\int _{1}^{4}{\frac {8}{\sqrt {x}}}dx&=&{\frac {8x^{1/2}}{2}}\vert _{1}^{4}\\&=&4x^{1/2}\vert _{1}^{4}\end{array}}}$
Step 3:
Now we need to evaluate to get:
${\displaystyle 4\cdot 4^{1/2}-4\cdot 1^{1/2}=8-4=4}$
Final Answer:
4