gamomaniea1

2021-11-20

Solve the definite integral.
${\int }_{1}^{2}\left(\frac{3{y}^{2}+y-1}{{y}^{2}}\right)$

### Answer & Explanation

Steven Arredondo

Step 1
We have to solve the given definite integral:
${\int }_{1}^{2}\left(\frac{3{y}^{2}+y-1}{{y}^{2}}\right)dy$
Rewriting the integral and solving the given integral,
${\int }_{1}^{2}\left(\frac{3{y}^{2}+y-1}{{y}^{2}}\right)dy={\int }_{1}^{2}\left(\frac{3{y}^{2}}{{y}^{2}}+\frac{y}{{y}^{2}}-\frac{1}{{y}^{2}}\right)dy$
$={\int }_{1}^{2}\left(3+\frac{1}{y}-{y}^{-2}\right)dy$
$={\left[3y+\mathrm{ln}\left(y\right)-\frac{{y}^{-2+1}}{-2+1}\right]}_{1}^{2}$
$={\left[3y+\mathrm{ln}\left(y\right)+{y}^{-1}\right]}_{1}^{2}$
$={\left[3y+\mathrm{ln}\left(y\right)+\frac{1}{y}\right]}_{1}^{2}$
$=\left[3×2+\mathrm{ln}\left(2\right)+\frac{1}{2}-3×1-\mathrm{ln}\left(1\right)-1\right]$
$=\left[6+\mathrm{ln}\left(2\right)+\frac{1}{2}-3-0-1\right]$
$=\left[2+\frac{1}{2}+\mathrm{ln}\left(2\right)\right]$
$=\frac{5}{2}+\mathrm{ln}\left(2\right)$
Step 2
Hence, value of given definite integral is $\frac{5}{2}+\mathrm{ln}\left(2\right)$.

Alicia Washington

Step 1: Simplify $\frac{3{y}^{2}+y-1}{{y}^{2}}\to \frac{1}{y}+\frac{3{y}^{2}-1}{{y}^{2}}$.
${\int }_{1}^{2}\frac{1}{y}+\frac{3{y}^{2}-1}{{y}^{2}}dy$
Step 2: 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 3: In this case, $f\left(y\right)=\frac{1}{y}+\frac{3{y}^{2}-1}{{y}^{2}}$. Find its integral.
$\mathrm{ln}y+3y+\frac{1}{y}{\mid }_{1}^{2}$
Step 4: Since $F\left(y\right){\mid }_{a}^{b}=F\left(b\right)-F\left(a\right)$, expand the above into F(2)−F(1):
$\left(\mathrm{ln}2+3×2+\frac{1}{2}\right)-\left(\mathrm{ln}1+3×1+\frac{1}{1}\right)$
Step 5: Simplify.
$\frac{5}{2}+\mathrm{ln}2$

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