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Popular >

integral from 1 to 2 of (x^3-x)

Solution

\int_{1}^{2} (x^{3} - x) d x

Solution

\frac{9}{4}

+1

Decimal Notation

2.25

Solution steps

\int_{1}^{2} x^{3} - x 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_{1}^{2} x^{3} d x - \int_{1}^{2} x d x

\int_{1}^{2} x^{3} d x = \frac{15}{4}

\int_{1}^{2} x d x = \frac{3}{2}

= \frac{15}{4} - \frac{3}{2}

Simplify

\frac{15}{4} - \frac{3}{2} : \frac{9}{4}

= \frac{9}{4}

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\int_{0}^{x} e^{- (x + y)} d y

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\int_{4}^{7} 1

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\int_{1}^{x} sin (x^{2}) d x

Frequently Asked Questions (FAQ)

What is the integral from 1 to 2 of (x^3-x) ?
The integral from 1 to 2 of (x^3-x) is 9/4