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

The integral from 0 to 1 of (1+x)^2 is 7/3

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

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

\frac{7}{3}

Decimal

2.33333 \dots

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

1 + 1 + \frac{1}{3} : \frac{7}{3}

= \frac{7}{3}

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What is the integral from 0 to 1 of (1+x)^2 ?
The integral from 0 to 1 of (1+x)^2 is 7/3