What is the integral from 0 to 1 of 4t^3sin(t^4) ?

The integral from 0 to 1 of 4t^3sin(t^4) is -cos(1)+1

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integral from 0 to 1 of 4t^3sin(t^4)

\int_{0}^{1} 4 t^{3} sin (t^{4}) d t

- cos (1) + 1

Decimal Notation

0.45969 \dots

Solution steps

\int_{0}^{1} 4 t^{3} sin (t^{4}) d t

Take the constant out:

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

= 4 \cdot \int_{0}^{1} t^{3} sin (t^{4}) d t

Apply u-substitution

: \int_{0}^{1} \frac{sin ( u )}{4} d u

= 4 \cdot \int_{0}^{1} \frac{sin ( u )}{4} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int_{0}^{1} sin (u) d u

Use the common integral:

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

= 4 \cdot \frac{1}{4} [- cos (u)]_{0}^{1}

Simplify

4 \cdot \frac{1}{4} [- cos (u)]_{0}^{1} : [- cos (u)]_{0}^{1}

= [- cos (u)]_{0}^{1}

Compute the boundaries:

- cos (1) + 1

= - cos (1) + 1

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What is the integral from 0 to 1 of 4t^3sin(t^4) ?
The integral from 0 to 1 of 4t^3sin(t^4) is -cos(1)+1