022 Exam 2 Sample A, Problem 6

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

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

 Solution:

Step 1:
Set up the integral:
$\int _{1}^{\,4}{\frac {8}{\sqrt {x}}}\,dx.$

Step 2:
Using the power rule we have:
${\begin{array}{rcl}\displaystyle {\int _{1}^{\,4}{\frac {8}{\sqrt {x}}}\,dx}&=&\displaystyle {\int _{1}^{\,4}8x^{-1/2}\,dx}\\\\&=&\displaystyle {{\frac {8x^{1/2}}{1/2}}{\Bigr |}_{x=1}^{4}}\\\\&=&16x^{1/2}{\Bigr |}_{x=1}^{4}.\end{array}}$

Step 3:
Now we need to evaluate to get:
$16x^{1/2}{\Bigr |}_{x=1}^{4}\,=\,16\cdot 4^{1/2}-16\cdot 1^{1/2}\,=\,32-16\,=\,16.$

Final Answer:
$\int _{1}^{\,4}{\frac {8}{\sqrt {x}}}\,dx\,=\,16.$

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