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Popular >

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

Solution

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

Solution

\frac{19}{3}

+1

Decimal

6.33333 \dots

Solution steps

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

Expand

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

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

\int_{1}^{2} x^{2} d x = \frac{7}{3}

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

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

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

Simplify

\frac{7}{3} + 3 + 1 : \frac{19}{3}

= \frac{19}{3}

Graph

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e^{5 (t - 2)} (t^{3} + 2 t)^{2}

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(4dx)/(dt)=4-2x

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integral of (x+3)^{10}

\int (x + 3)^{10} d x

inverse oflaplace s^4 inverse laplace

s^{4}

Frequently Asked Questions (FAQ)

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