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Popular Calculus >

integral from-1 to 1 of (4x^2+1)

Solution

\int_{- 1}^{1} (4 x^{2} + 1) d x

Solution

\frac{14}{3}

+1

Decimal

4.66666 \dots

Solution steps

\int_{- 1}^{1} 4 x^{2} + 1 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}^{1} 4 x^{2} d x + \int_{- 1}^{1} 1 d x

\int_{- 1}^{1} 4 x^{2} d x = \frac{8}{3}

\int_{- 1}^{1} 1 d x = 2

= \frac{8}{3} + 2

Simplify

\frac{8}{3} + 2 : \frac{14}{3}

= \frac{14}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the integral from-1 to 1 of (4x^2+1) ?
The integral from-1 to 1 of (4x^2+1) is 14/3