What is the integral from 0 to 1 of xcos(2x^2+7) ?

The integral from 0 to 1 of xcos(2x^2+7) is 1/4 (sin(9)-sin(7))

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integral from 0 to 1 of xcos(2x^2+7)

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

\frac{1}{4} (sin (9) - sin (7))

Decimal Notation

- 0.06121 \dots

Solution steps

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

Apply u-substitution

: \int_{7}^{9} \frac{cos ( u )}{4} d u

= \int_{7}^{9} \frac{cos ( u )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{7}^{9} cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{4} [sin (u)]_{7}^{9}

Compute the boundaries:

sin (9) - sin (7)

= \frac{1}{4} (sin (9) - sin (7))

\frac{\partial}{\partial x} (x y^{2} + y^{2})

\int_{1}^{3} 2 r^{4} ln (r) d r

(- 4, - 4), (2, 5)

\int (4 x - 3) (2 x - 1) d x

\frac{\partial}{\partial x} (x^{3} y + x y^{2} - 2 x^{2} - y^{2})

What is the integral from 0 to 1 of xcos(2x^2+7) ?
The integral from 0 to 1 of xcos(2x^2+7) is 1/4 (sin(9)-sin(7))