What is the integral from 3 to 4 of (-x^2+17) ?

The integral from 3 to 4 of (-x^2+17) is 14/3

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integral from 3 to 4 of (-x^2+17)

\int_{3}^{4} (- x^{2} + 17) d x

\frac{14}{3}

Decimal

4.66666 \dots

Solution steps

\int_{3}^{4} - x^{2} + 17 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_{3}^{4} x^{2} d x + \int_{3}^{4} 17 d x

\int_{3}^{4} x^{2} d x = \frac{37}{3}

\int_{3}^{4} 17 d x = 17

= - \frac{37}{3} + 17

Simplify

- \frac{37}{3} + 17 : \frac{14}{3}

= \frac{14}{3}

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What is the integral from 3 to 4 of (-x^2+17) ?
The integral from 3 to 4 of (-x^2+17) is 14/3