What is the integral from 1 to 7 of (6x^2+2x-3) ?

The integral from 1 to 7 of (6x^2+2x-3) is 714

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integral from 1 to 7 of (6x^2+2x-3)

\int_{1}^{7} (6 x^{2} + 2 x - 3) d x

714

Solution steps

\int_{1}^{7} 6 x^{2} + 2 x - 3 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}^{7} 6 x^{2} d x + \int_{1}^{7} 2 x d x - \int_{1}^{7} 3 d x

\int_{1}^{7} 6 x^{2} d x = 684

\int_{1}^{7} 2 x d x = 48

\int_{1}^{7} 3 d x = 18

= 684 + 48 - 18

Simplify

= 714

n = 10 \sum \infty \frac{2}{3} (\frac{3}{4})^{n}

\int (1 + x)^{8} d x

\frac{7 x}{x - 1}

\int_{0}^{\infty} x e^{- x y} d x

(x) (3 - x)^{2}

What is the integral from 1 to 7 of (6x^2+2x-3) ?
The integral from 1 to 7 of (6x^2+2x-3) is 714