Ashley Searcy

2021-11-18

Evaluate the integral.
${\int }_{-1}^{1}{\left(x-3\right)}^{2}dx$

Fesion

Step 1
Given :
${\int }_{-1}^{1}{\left(x-3\right)}^{2}dx$
To find the value of the above integral.
Step 2
${\int }_{-1}^{1}{\left(x-3\right)}^{2}dx$
$={\int }_{-1}^{1}\left({x}^{2}-6x+9\right)dx$ (since ${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$)
$={\left[\frac{{x}^{3}}{3}-6\frac{{x}^{2}}{2}+9x\right]}_{-1}^{1}$
$=\left(\frac{{1}^{3}}{3}-6\frac{{1}^{2}}{2}+9\left(1\right)\right)-\left(\frac{{\left(-1\right)}^{3}}{3}-6\frac{{\left(-1\right)}^{2}}{2}+9\left(-1\right)\right)$
$=\left(\frac{1}{3}-3+9\right)-\left(\frac{-1}{3}-3-9\right)$
$=\left(\frac{1}{3}+6\right)-\left(\frac{-1}{3}-12\right)$
$=\left(\frac{1+18}{3}\right)-\left(\frac{-1-36}{3}\right)=\frac{19}{3}+\frac{37}{3}=\frac{56}{3}$
Thus, ${\int }_{-1}^{1}{\left(x-3\right)}^{2}dx=\frac{56}{3}$.

Geraldine Flores

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, $f\left(x\right)={\left(x-3\right)}^{2}$. Find its integral.
$\frac{{x}^{3}}{3}-3{x}^{2}+9x{\mid }_{-1}^{1}$
Step 3: Since $F\left(x\right){\mid }_{a}^{b}=F\left(b\right)-F\left(a\right)$, expand the above into F(1)−F(−1):
$\left(\frac{{1}^{3}}{3}-3×{1}^{2}+9×1\right)-\left(\frac{{\left(-1\right)}^{3}}{3}-3{\left(-1\right)}^{2}+9×-1\right)$
Step 4: Simplify.
$\frac{56}{3}$

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