What is the integral from-1 to 2 of 4x^3 ?

The integral from-1 to 2 of 4x^3 is 15

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integral from-1 to 2 of 4x^3

\int_{- 1}^{2} 4 x^{3} d x

15

Solution steps

\int_{- 1}^{2} 4 x^{3} d x

Take the constant out:

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

= 4 \cdot \int_{- 1}^{2} x^{3} d x

Apply the Power Rule

: [\frac{x ^{4}}{4}]_{- 1}^{2}

= 4 [\frac{x ^{4}}{4}]_{- 1}^{2}

Compute the boundaries:

\frac{15}{4}

= 4 \cdot \frac{15}{4}

Simplify

= 15

\int_{- π}^{π} sin^{2} (2 x) d x

\int_{1}^{2} \frac{1}{x ^{2} + 1} d x

\int_{0}^{4} 2 x + 1 d x

\int_{0}^{\frac{π}{2}} 2 cos^{2} (θ) d θ

\int_{0}^{2} 1 - x^{2} d x

What is the integral from-1 to 2 of 4x^3 ?
The integral from-1 to 2 of 4x^3 is 15