What is the integral from-2 to 4 of x^2sin(x^3+3) ?

The integral from-2 to 4 of x^2sin(x^3+3) is 1/3 (-cos(67)+cos(5))

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integral from-2 to 4 of x^2sin(x^3+3)

\int_{- 2}^{4} x^{2} sin (x^{3} + 3) d x

\frac{1}{3} (- cos (67) + cos (5))

Decimal Notation

0.26714 \dots

Solution steps

\int_{- 2}^{4} x^{2} sin (x^{3} + 3) d x

Apply u-substitution

: \int_{- 8}^{64} \frac{sin ( u + 3 )}{3} d u

= \int_{- 8}^{64} \frac{sin ( u + 3 )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{- 8}^{64} sin (u + 3) d u

Apply u-substitution

: \int_{- 5}^{67} sin (v) d v

= \frac{1}{3} \cdot \int_{- 5}^{67} sin (v) d v

Use the common integral:

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

= \frac{1}{3} [- cos (v)]_{- 5}^{67}

Compute the boundaries:

- cos (67) + cos (5)

= \frac{1}{3} (- cos (67) + cos (5))

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What is the integral from-2 to 4 of x^2sin(x^3+3) ?
The integral from-2 to 4 of x^2sin(x^3+3) is 1/3 (-cos(67)+cos(5))