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Popular Calculus >

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

Solution

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

Solution

\frac{19}{3}

+1

Decimal

6.33333 \dots

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

\frac{1}{3} + 2 + 4 : \frac{19}{3}

= \frac{19}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the integral from 0 to 1 of (x+2)^2 ?
The integral from 0 to 1 of (x+2)^2 is 19/3