022 Exam 2 Sample A, Problem 6

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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:
$16x^{1/2}{\Bigr |}_{x=1}^{4}\,=\,16\cdot 4^{1/2}-16\cdot 1^{1/2}\,=\,32-16\,=\,16.$ $\int _{1}^{\,4}{\frac {8}{\sqrt {x}}}\,dx\,=\,16.$ 