What is the integral from 0 to pi of 7x^2sin(1/2 x) ?

The integral from 0 to pi of 7x^2sin(1/2 x) is 7(8pi-16)

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integral from 0 to pi of 7x^2sin(1/2 x)

\int_{0}^{π} 7 x^{2} sin (\frac{1}{2} x) d x

7 (8 π - 16)

Decimal Notation

63.92918 \dots

Solution steps

\int_{0}^{π} 7 x^{2} sin (\frac{1}{2} x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int_{0}^{π} x^{2} sin (\frac{1}{2} x) d x

Simplify

\frac{1}{2} x : \frac{x}{2}

= 7 \cdot \int_{0}^{π} x^{2} sin (\frac{x}{2}) d x

Apply Integration By Parts

: [- 2 x^{2} cos (\frac{x}{2}) - \int - 4 x cos (\frac{x}{2}) d x]_{0}^{π}

= 7 [- 2 x^{2} cos (\frac{x}{2}) - \int - 4 x cos (\frac{x}{2}) d x]_{0}^{π}

\int - 4 x cos (\frac{x}{2}) d x = - 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))

= 7 [- 2 x^{2} cos (\frac{x}{2}) - (- 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2})))]_{0}^{π}

Simplify

= 7 [- 2 x^{2} cos (\frac{x}{2}) + 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))]_{0}^{π}

Compute the boundaries:

8 π - 16

= 7 (8 π - 16)

\frac{1}{( 1 - x )}, 0

x

y = x^{4} + 5 x^{2} - x

f (x) = 6 x^{2}

What is the integral from 0 to pi of 7x^2sin(1/2 x) ?
The integral from 0 to pi of 7x^2sin(1/2 x) is 7(8pi-16)