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Popular >

integral from 0 to 1 of (x^3+4x)

Solution

\int_{0}^{1} (x^{3} + 4 x) d x

Solution

\frac{9}{4}

+1

Solution steps

\int_{0}^{1} x^{3} + 4 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_{0}^{1} x^{3} d x + \int_{0}^{1} 4 x d x

\int_{0}^{1} x^{3} d x = \frac{1}{4}

\int_{0}^{1} 4 x d x = 2

= \frac{1}{4} + 2

Simplify

\frac{1}{4} + 2 : \frac{9}{4}

= \frac{9}{4}

Popular Examples

integral from 1 to 2 of 1/(y+4)

\int_{1}^{2} \frac{1}{y + 4} d y

integral from 0 to k of 8^{x/k}

\int_{0}^{k} 8^{\frac{x}{k}} d x

integral from 1 to 2 of 3x-x^2+2

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

integral from 0 to 1 of sqrt(2+1/(4x))

\int_{0}^{1} 2 + \frac{1}{4 x} d x

integral from 0 to 1 of cos(θ)

\int_{0}^{1} cos (θ) d θ

Frequently Asked Questions (FAQ)

What is the integral from 0 to 1 of (x^3+4x) ?
The integral from 0 to 1 of (x^3+4x) is 9/4