What is the integral from 1 to 3 of (x^2-4x+8) ?

The integral from 1 to 3 of (x^2-4x+8) is 26/3

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integral from 1 to 3 of (x^2-4x+8)

\int_{1}^{3} (x^{2} - 4 x + 8) d x

\frac{26}{3}

Decimal

8.66666 \dots

Solution steps

\int_{1}^{3} x^{2} - 4 x + 8 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}^{3} x^{2} d x - \int_{1}^{3} 4 x d x + \int_{1}^{3} 8 d x

\int_{1}^{3} x^{2} d x = \frac{26}{3}

\int_{1}^{3} 4 x d x = 16

\int_{1}^{3} 8 d x = 16

= \frac{26}{3} - 16 + 16

Simplify

= \frac{26}{3}

\int_{0}^{1} (- x) d x

\int_{0}^{\frac{π}{2}} x sin^{2} (\frac{x}{2}) d x

\int_{0}^{4 π} sin^{2} (x) d x

\int_{0}^{\infty} \frac{3}{x ^{2}} d x

\int_{2}^{\infty} \frac{3}{x ^{2}} d x

What is the integral from 1 to 3 of (x^2-4x+8) ?
The integral from 1 to 3 of (x^2-4x+8) is 26/3