# Difference between revisions of "022 Exam 2 Sample A, Problem 6"

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

Foundations:
For solving the problem, we only require the use of the power rule for integration:
${\displaystyle \int x^{n}\,dx\,=\,{\frac {x^{n+1}}{n+1}}+C,}$  for ${\displaystyle n\neq -1.}$
Geometrically, we need to integrate the region between the ${\displaystyle x}$-axis, the curve, and the vertical lines ${\displaystyle x=1}$ and ${\displaystyle x=4}$.

Solution:

Step 1:
Set up the integral:
${\displaystyle \int _{1}^{\,4}{\frac {8}{\sqrt {x}}}\,dx.}$
Step 2:
Using the power rule we have:
${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{1}^{\,4}{\frac {8}{\sqrt {x}}}\,dx}&=&\displaystyle {\int _{1}^{\,4}8x^{-1/2}\,dx}\\\\&=&\displaystyle {{\frac {8x^{1/2}}{1/2}}{\Bigr |}_{x=1}^{4}}\\\\&=&16x^{1/2}{\Bigr |}_{x=1}^{4}.\end{array}}}$
Step 3:
Now we need to evaluate to get:
${\displaystyle 16x^{1/2}{\Bigr |}_{x=1}^{4}\,=\,16\cdot 4^{1/2}-16\cdot 1^{1/2}\,=\,32-16\,=\,16.}$
${\displaystyle \int _{1}^{\,4}{\frac {8}{\sqrt {x}}}\,dx\,=\,16.}$