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Popular >

integral from 0 to 3 of |x^2-9|

Solution

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

Solution

18

Solution steps

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

Eliminate Absolutes

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int_{0}^{3} 9 d x = 27

= - 9 + 27

Simplify

= 18

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\int - e^{- 2 x} d x

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\frac{64}{x ^{2} + 16}

tangent of f(x)=x^2-4x,\at x=1 tangent of

f (x) = x^{2} - 4 x, at x = 1

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\frac{d}{d x} (- 5 sin (x))

(\partial)/(\partial x)(sqrt(2-x^2-y^2))

\frac{\partial}{\partial x} (2 - x^{2} - y^{2})

Frequently Asked Questions (FAQ)

What is the integral from 0 to 3 of |x^2-9| ?
The integral from 0 to 3 of |x^2-9| is 18