What is the integral from 1 to 2 of xsin(x^2) ?

The integral from 1 to 2 of xsin(x^2) is 1/2 (-cos(4)+cos(1))

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integral from 1 to 2 of xsin(x^2)

\int_{1}^{2} x sin (x^{2}) d x

\frac{1}{2} (- cos (4) + cos (1))

Decimal Notation

0.59697 \dots

Solution steps

\int_{1}^{2} x sin (x^{2}) d x

Apply u-substitution

: \int_{1}^{4} \frac{sin ( u )}{2} d u

= \int_{1}^{4} \frac{sin ( u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{4} sin (u) d u

Use the common integral:

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

= \frac{1}{2} [- cos (u)]_{1}^{4}

Compute the boundaries:

- cos (4) + cos (1)

= \frac{1}{2} (- cos (4) + cos (1))

\frac{d}{d x} (\frac{72}{x})

\int_{0}^{π} (cos (2 x))^{4} d x

\int 3 x \cdot e^{(- 4 x)} d x

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

y = x^{2} - 8

What is the integral from 1 to 2 of xsin(x^2) ?
The integral from 1 to 2 of xsin(x^2) is 1/2 (-cos(4)+cos(1))