What is the integral from 3 to 4 of (x^2+x-12) ?

The integral from 3 to 4 of (x^2+x-12) is 23/6

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integral from 3 to 4 of (x^2+x-12)

\int_{3}^{4} (x^{2} + x - 12) d x

\frac{23}{6}

Decimal

3.83333 \dots

Solution steps

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

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

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

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

= \frac{37}{3} + \frac{7}{2} - 12

Simplify

\frac{37}{3} + \frac{7}{2} - 12 : \frac{23}{6}

= \frac{23}{6}

\int_{1}^{2} x + 6 d x

\int_{- 2}^{4} 4 d x

\int_{0}^{2} 4 x e^{- 2 x} d x

\int_{0}^{1} (\frac{5}{7 + 7 x ^{2}}) d x

\int_{0}^{2} x x d x

What is the integral from 3 to 4 of (x^2+x-12) ?
The integral from 3 to 4 of (x^2+x-12) is 23/6