# Evaluate the integral. \int_{2}^{3}[3x^{2}+2x+\frac{1}{x^{2}}]dx

Evaluate the integral.
${\int }_{2}^{3}\left[3{x}^{2}+2x+\frac{1}{{x}^{2}}\right]dx$
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breisgaoyz
Step 1
The given integral is ${\int }_{2}^{3}\left[3{x}^{2}+2x+\frac{1}{{x}^{2}}\right]dx$.
Step 2
Evaluate the given integral as shown below:
${\int }_{2}^{3}\left[3{x}^{2}+2x+\frac{1}{{x}^{2}}\right]dx={\int }_{2}^{3}\left[3{x}^{2}\right]dx+{\int }_{2}^{3}\left[2x\right]dx+{\int }_{2}^{3}\left[\frac{1}{{x}^{2}}\right]dx$
$=3{\int }_{2}^{3}{x}^{2}dx+2{\int }_{2}^{3}xdx+{\int }_{2}^{3}{x}^{-2}dx$
$=3{\left[\frac{{x}^{3}}{3}\right]}_{2}^{3}+2{\left[\frac{{x}^{2}}{2}\right]}_{2}^{3}+{\left[\frac{{x}^{-1}}{-1}\right]}_{2}^{3}$
$={\left[{x}^{3}\right]}_{2}^{3}+{\left[{x}^{2}\right]}_{2}^{3}-{\left[\frac{1}{x}\right]}_{2}^{3}$
$=\left[27-8\right]+\left[9-4\right]-\left[\frac{1}{3}-\frac{1}{2}\right]$
$=\left[19\right]+\left[5\right]+\left[\frac{1}{6}\right]$
$=\frac{145}{6}$
###### Not exactly what you’re looking for?
Rosemary McBride
Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:
${\int }_{a}^{b}f\left(x\right)dx=F\left(x\right){\mid }_{a}^{b}=F\left(b\right)-F\left(a\right)$
Step 2: In this case, $f\left(x\right)=\left(3{x}^{2}+2x+\frac{1}{{x}^{2}}\right)$. Find its integral.
${x}^{3}+{x}^{2}-\frac{1}{x}{\mid }_{2}^{3}$
Step 3: Since $F\left(x\right){\mid }_{a}^{b}=F\left(b\right)-F\left(a\right)$, expand the above into F(3)-F(2):
$\left({3}^{3}+{3}^{2}-\frac{1}{3}\right)-\left({2}^{3}+{2}^{2}-\frac{1}{2}\right)$
Step 4: Simplify.
$\frac{145}{6}$