Question

The value of integral $\int |2x^2-8|dx$ will be

• A. $8x+\frac{2x^3}{3}+C$ for $x\in [0,2]$
• B. $\frac{2x^3}{3}-8x+C$ for $x\in (2,3)$
• C. $8x-\frac{2x^3}{3}+C$ for $x\in (0,2)$
• D. $\frac{2x^3}{3}+8x+C$ for $x\in [2,3]$

Answer 1

Answer: (B), (C)

$\frac{2x^3}{3}-8x+C$ for $x\in (2,3)$
$8x-\frac{2x^3}{3}+C$ for $x\in (0,2)$

To find: $\int |2x^2-8|dx$

Now, roots of $2x^2-8$ are $+2$ and $-2$

The function is +ve in the interval $x\in (-\infty ,-2)\cup (2,\infty )$

The function is -ve in $x\in (-2,2)$

$\int |2x^2-8|dx$

$=\int -(2x^2-8)dxforx\in (-2,2)$

$=8x-\frac{2x^3}{3}$

$=\int (2x^2-8)forx\in (-2,\infty )\cup (2,\infty )$

$=\frac{2x^3}{3}-8x$