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

The integral from 0 to pi/2 of 10+sin(x)-x^2 is 5pi+1-(pi^3)/(24)

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integral from 0 to pi/2 of 10+sin(x)-x^2

\int_{0}^{\frac{π}{2}} 10 + sin (x) - x^{2} d x

5 π + 1 - \frac{π ^{3}}{24}

Decimal

15.41603 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 10 + sin (x) - x^{2} 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}^{\frac{π}{2}} 10 d x + \int_{0}^{\frac{π}{2}} sin (x) d x - \int_{0}^{\frac{π}{2}} x^{2} d x

\int_{0}^{\frac{π}{2}} 10 d x = 5 π

\int_{0}^{\frac{π}{2}} sin (x) d x = 1

\int_{0}^{\frac{π}{2}} x^{2} d x = \frac{π ^{3}}{24}

= 5 π + 1 - \frac{π ^{3}}{24}

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What is the integral from 0 to pi/2 of 10+sin(x)-x^2 ?
The integral from 0 to pi/2 of 10+sin(x)-x^2 is 5pi+1-(pi^3)/(24)