What is the integral from 1 to 3 of 2xcos(x^2) ?

The integral from 1 to 3 of 2xcos(x^2) is sin(9)-sin(1)

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integral from 1 to 3 of 2xcos(x^2)

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

sin (9) - sin (1)

Decimal

- 0.42935 \dots

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{1}^{3} x cos (x^{2}) d x

Apply u-substitution

: \int_{1}^{9} \frac{cos ( u )}{2} d u

= 2 \cdot \int_{1}^{9} \frac{cos ( u )}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{1}^{9} cos (u) d u

Use the common integral:

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

= 2 \cdot \frac{1}{2} [sin (u)]_{1}^{9}

Simplify

2 \cdot \frac{1}{2} [sin (u)]_{1}^{9} : [sin (u)]_{1}^{9}

= [sin (u)]_{1}^{9}

Compute the boundaries:

sin (9) - sin (1)

= sin (9) - sin (1)

\int_{2}^{5} \frac{10}{9 x ^{2}} d x

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

\int_{1}^{3} \frac{x}{x ^{2} + x - 6} d x

\int_{- 1}^{1} x^{3} + x^{5} + 7 x^{9} d x

\int_{0}^{16} x - 3 d x

What is the integral from 1 to 3 of 2xcos(x^2) ?
The integral from 1 to 3 of 2xcos(x^2) is sin(9)-sin(1)