What is the integral from 0 to pi of (3x^2y-4sin(y)) ?

The integral from 0 to pi of (3x^2y-4sin(y)) is -8+(3pi^2x^2)/2

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integral from 0 to pi of (3x^2y-4sin(y))

\int_{0}^{π} (3 x^{2} y - 4 sin (y)) d y

- 8 + \frac{3 π ^{2} x ^{2}}{2}

Solution steps

\int_{0}^{π} 3 x^{2} y - 4 sin (y) d y

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} 4 sin (y) d y + \int_{0}^{π} 3 x^{2} y d y

\int_{0}^{π} 4 sin (y) d y = 8

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

= - 8 + \frac{3 π ^{2} x ^{2}}{2}

\int_{- 2}^{- 1} (x^{- 4}) d x

\int_{- 1}^{2} (3 x^{2} - 5 x)^{3} d x

\int_{0.75}^{1} 2 x d x

\int_{- 2}^{1} (2 x^{2} + 4) d x

\int_{0}^{4} x^{2} - 4 x + 9 d x

What is the integral from 0 to pi of (3x^2y-4sin(y)) ?
The integral from 0 to pi of (3x^2y-4sin(y)) is -8+(3pi^2x^2)/2