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Popular Calculus >

integral from 0 to 1 of (2x+1)^2

Solution

\int_{0}^{1} (2 x + 1)^{2} d x

Solution

\frac{13}{3}

+1

Decimal

4.33333 \dots

Solution steps

\int_{0}^{1} (2 x + 1)^{2} d x

Expand

(2 x + 1)^{2} : 4 x^{2} + 4 x + 1

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

\int_{0}^{1} 4 x^{2} d x = \frac{4}{3}

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

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

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

Simplify

\frac{4}{3} + 2 + 1 : \frac{13}{3}

= \frac{13}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the integral from 0 to 1 of (2x+1)^2 ?
The integral from 0 to 1 of (2x+1)^2 is 13/3