Integral Calculus> options -3/2 9/8 ¼ 3/2...

options

-3/2

9/8

¼

3/2

Aditya Kartikeya , 10 Years ago

Grade 10

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

$I = \frac{1}{2}\int_{1/2}^{7/2}[|x-3|+|x-1|-4]dx$
$I = \frac{1}{2}[\int_{1/2}^{1}[-x+3-x+1-4]dx+\int_{1}^{7/2}[|x-3|+|x-1|-4]dx]$ $I = \frac{1}{2}[\int_{1/2}^{1}(-2x)dx+\int_{1}^{7/2}[|x-3|+|x-1|-4]dx]$
$I = \frac{1}{2}[-(x^{2})_{1/2}^{1}+\int_{1}^{7/2}[|x-3|+|x-1|-4]dx]$
$I = \frac{1}{2}[\frac{-3}{4}+\int_{1}^{7/2}[|x-3|+|x-1|-4]dx]$
$I = \frac{1}{2}[\frac{-3}{4}+\int_{1}^{3}[-x+3+x-1-4]dx+\int_{3}^{7/2}[x-3+x-1-4]dx]$ $I = \frac{1}{2}[\frac{-3}{4}+\int_{1}^{3}[-2]dx+\int_{3}^{7/2}[2x-8]dx]$
$I = \frac{1}{2}[\frac{-3}{4}+[-2x]_{1}^{3}+[x^2-8x]_{3}^{7/2}]$
$I = \frac{1}{2}[\frac{-3}{4}-4+[(\frac{49}{4}-28)-(9-24)]]$
$I = \frac{1}{2}[\frac{-3}{4}-4+[(\frac{49}{4}-15)]]$
$I = \frac{1}{2}[\frac{-3}{4}-4-\frac{11}{4}]$
$I = \frac{1}{2}[-4-\frac{7}{2}]$