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Popular >

integral from 0 to 4 of |x^2-9|

Solution

\int_{0}^{4} x^{2} - 9 d x

Solution

\frac{64}{3}

+1

Solution steps

\int_{0}^{4} x^{2} - 9 d x

Eliminate Absolutes

= \int_{0}^{3} - x^{2} + 9 d x + \int_{3}^{4} x^{2} - 9 d x

\int_{0}^{3} - x^{2} + 9 d x = 18

\int_{3}^{4} x^{2} - 9 d x = \frac{10}{3}

= 18 + \frac{10}{3}

Simplify

18 + \frac{10}{3} : \frac{64}{3}

= \frac{64}{3}

Popular Examples

inverse oflaplace 3/(2s+1)inverse laplace

\frac{3}{2 s + 1}

tangent of f(x)=14-e^x tangent of

f (x) = 14 - e^{x}

derivative of f(x)=2sqrt(x)+1 derivative of

f (x) = 2 x + 1

derivative of (2x-7)^3 derivative of

(2 x - 7)^{3}

limit as x approaches 0 of (sec^2(3x))/2

x \to 0 lim (\frac{sec ^{2} ( 3 x )}{2})

Frequently Asked Questions (FAQ)

What is the integral from 0 to 4 of |x^2-9| ?
The integral from 0 to 4 of |x^2-9| is 64/3