What is the integral from 0 to pi of (10+x) ?

The integral from 0 to pi of (10+x) is 10pi+(pi^2)/2

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integral from 0 to pi of (10+x)

\int_{0}^{π} (10 + x) d x

10 π + \frac{π ^{2}}{2}

Decimal

36.35072 \dots

Solution steps

\int_{0}^{π} 10 + x 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}^{π} 10 d x + \int_{0}^{π} x d x

\int_{0}^{π} 10 d x = 10 π

\int_{0}^{π} x d x = \frac{π ^{2}}{2}

= 10 π + \frac{π ^{2}}{2}

\int_{4}^{3 x + 4} \frac{t}{1 + t ^{3}} d t

\int_{0}^{π} e^{2 x} d x

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

\int_{1}^{8} (\frac{x}{2} - \frac{9}{x}) d x

What is the integral from 0 to pi of (10+x) ?
The integral from 0 to pi of (10+x) is 10pi+(pi^2)/2