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options -3/2 9/8 ¼ 3/2

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  1. -3/2
  2. 9/8
  3. ¼
  4. 3/2

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Grade:10

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 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}]

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