Ashley Searcy

2021-11-18

Evaluate the integral.

${\int}_{-1}^{1}{(x-3)}^{2}dx$

Fesion

Beginner2021-11-19Added 24 answers

Step 1

Given :

${\int}_{-1}^{1}{(x-3)}^{2}dx$

To find the value of the above integral.

Step 2

${\int}_{-1}^{1}{(x-3)}^{2}dx$

$={\int}_{-1}^{1}({x}^{2}-6x+9)dx$ (since $(a-b)}^{2}={a}^{2}-2ab+{b}^{2$ )

$={[\frac{{x}^{3}}{3}-6\frac{{x}^{2}}{2}+9x]}_{-1}^{1}$

$=(\frac{{1}^{3}}{3}-6\frac{{1}^{2}}{2}+9\left(1\right))-(\frac{{(-1)}^{3}}{3}-6\frac{{(-1)}^{2}}{2}+9(-1))$

$=(\frac{1}{3}-3+9)-(\frac{-1}{3}-3-9)$

$=(\frac{1}{3}+6)-(\frac{-1}{3}-12)$

$=\left(\frac{1+18}{3}\right)-\left(\frac{-1-36}{3}\right)=\frac{19}{3}+\frac{37}{3}=\frac{56}{3}$

Thus,$\int}_{-1}^{1}{(x-3)}^{2}dx=\frac{56}{3$ .

Given :

To find the value of the above integral.

Step 2

Thus,

Geraldine Flores

Beginner2021-11-20Added 21 answers

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,$f\left(x\right)={(x-3)}^{2}$ . 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):

$(\frac{{1}^{3}}{3}-3\times {1}^{2}+9\times 1)-(\frac{{(-1)}^{3}}{3}-3{(-1)}^{2}+9\times -1)$

Step 4: Simplify.

$\frac{56}{3}$

Step 2: In this case,

Step 3: Since

Step 4: Simplify.

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