What is the integral from 0 to 4 of (7x^2-8x+7) ?

The integral from 0 to 4 of (7x^2-8x+7) is 340/3

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integral from 0 to 4 of (7x^2-8x+7)

\int_{0}^{4} (7 x^{2} - 8 x + 7) d x

\frac{340}{3}

Decimal

113.33333 \dots

Solution steps

\int_{0}^{4} 7 x^{2} - 8 x + 7 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_{0}^{4} 7 x^{2} d x - \int_{0}^{4} 8 x d x + \int_{0}^{4} 7 d x

\int_{0}^{4} 7 x^{2} d x = \frac{448}{3}

\int_{0}^{4} 8 x d x = 64

\int_{0}^{4} 7 d x = 28

= \frac{448}{3} - 64 + 28

Simplify

\frac{448}{3} - 64 + 28 : \frac{340}{3}

= \frac{340}{3}

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What is the integral from 0 to 4 of (7x^2-8x+7) ?
The integral from 0 to 4 of (7x^2-8x+7) is 340/3