Aspire Faculty ID #17240 · Topic: CUET UG 2022 Applied Mathematics · Just now
CUET UG 2022 Applied Mathematics

$\int_{-1}^{1}(|x-2|+|x|),dx=$

Solution

On interval $[-1,1]$, we have $x-2<0$

So,

$|x-2|=2-x$

Now split $|x|$ at $x=0$

For $-1\le x\le 0$

$|x|=-x$

Integrand:

$(2-x-x)=2-2x$

For $0\le x\le 1$

$|x|=x$

Integrand:

$(2-x+x)=2$

Now integrate:

$\int_{-1}^{0}(2-2x),dx=[2x-x^2]_{-1}^{0}$

$=0-(-3)=3$

Next,

$\int_{0}^{1}2,dx=2$

Total value:

$3+2=5$
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