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Popular Calculus >

integral from 1 to 2 of (2x-3)^2

Solution

\int_{1}^{2} (2 x - 3)^{2} d x

Solution

\frac{1}{3}

+1

Decimal

0.33333 \dots

Solution steps

\int_{1}^{2} (2 x - 3)^{2} d x

Expand

(2 x - 3)^{2} : 4 x^{2} - 12 x + 9

= \int_{1}^{2} 4 x^{2} - 12 x + 9 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} 4 x^{2} d x - \int_{1}^{2} 12 x d x + \int_{1}^{2} 9 d x

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

\int_{1}^{2} 12 x d x = 18

\int_{1}^{2} 9 d x = 9

= \frac{28}{3} - 18 + 9

Simplify

\frac{28}{3} - 18 + 9 : \frac{1}{3}

= \frac{1}{3}

Graph

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Frequently Asked Questions (FAQ)

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