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Popular Calculus >

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

Solution

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

Solution

\frac{9}{4}

+1

Decimal

2.25

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}

Graph

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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