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Popular Calculus >

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

Solution

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

Solution

\frac{68}{3}

+1

Decimal

22.66666 \dots

Solution steps

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

\int_{- 1}^{3} 2 x^{2} d x = \frac{56}{3}

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

= \frac{56}{3} + 4

Simplify

\frac{56}{3} + 4 : \frac{68}{3}

= \frac{68}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the integral from-1 to 3 of (2x^2+1) ?
The integral from-1 to 3 of (2x^2+1) is 68/3