What is the integral from 1 to 2 of x^3cos(x^4) ?

The integral from 1 to 2 of x^3cos(x^4) is 1/4 (sin(16)-sin(1))

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integral from 1 to 2 of x^3cos(x^4)

\int_{1}^{2} x^{3} cos (x^{4}) d x

\frac{1}{4} (sin (16) - sin (1))

Decimal Notation

- 0.28234 \dots

Solution steps

\int_{1}^{2} x^{3} cos (x^{4}) d x

Apply u-substitution

: \int_{1}^{16} \frac{cos ( u )}{4} d u

= \int_{1}^{16} \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_{1}^{16} cos (u) d u

Use the common integral:

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

= \frac{1}{4} [sin (u)]_{1}^{16}

Compute the boundaries:

sin (16) - sin (1)

= \frac{1}{4} (sin (16) - sin (1))

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What is the integral from 1 to 2 of x^3cos(x^4) ?
The integral from 1 to 2 of x^3cos(x^4) is 1/4 (sin(16)-sin(1))