What is the integral from 3 to 4 of (5-x^2)-(17-7x) ?

The integral from 3 to 4 of (5-x^2)-(17-7x) is 1/6

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integral from 3 to 4 of (5-x^2)-(17-7x)

\int_{3}^{4} (5 - x^{2}) - (17 - 7 x) d x

\frac{1}{6}

Decimal

0.16666 \dots

Solution steps

\int_{3}^{4} 5 - x^{2} - (17 - 7 x) d x

Simplify

= \int_{3}^{4} - x^{2} + 7 x - 12 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} 7 x d x - \int_{3}^{4} 12 d x

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

\int_{3}^{4} 7 x d x = \frac{49}{2}

\int_{3}^{4} 12 d x = 12

= - \frac{37}{3} + \frac{49}{2} - 12

Simplify

- \frac{37}{3} + \frac{49}{2} - 12 : \frac{1}{6}

= \frac{1}{6}

What is the integral from 3 to 4 of (5-x^2)-(17-7x) ?
The integral from 3 to 4 of (5-x^2)-(17-7x) is 1/6