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

The integral from 0 to 1 of (x^3+3x+4) is 23/4

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

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

\frac{23}{4}

Decimal

5.75

Solution steps

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

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

\int_{0}^{1} 3 x d x = \frac{3}{2}

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

= \frac{1}{4} + \frac{3}{2} + 4

Simplify

\frac{1}{4} + \frac{3}{2} + 4 : \frac{23}{4}

= \frac{23}{4}

What is the integral from 0 to 1 of (x^3+3x+4) ?
The integral from 0 to 1 of (x^3+3x+4) is 23/4