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Popular Calculus >

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

Solution

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

Solution

\frac{49}{3}

+1

Decimal

16.33333 \dots

Solution steps

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

Expand

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

= \int_{1}^{2} 1 + 4 y + 4 y^{2} d y

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} 1 d y + \int_{1}^{2} 4 y d y + \int_{1}^{2} 4 y^{2} d y

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

\int_{1}^{2} 4 y d y = 6

\int_{1}^{2} 4 y^{2} d y = \frac{28}{3}

= 1 + 6 + \frac{28}{3}

Simplify

1 + 6 + \frac{28}{3} : \frac{49}{3}

= \frac{49}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the integral from 1 to 2 of (1+2y)^2 ?
The integral from 1 to 2 of (1+2y)^2 is 49/3