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

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

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integral from-1 to 2 of (x^3)/3

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

\frac{5}{4}

Decimal Notation

1.25

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

= \frac{1}{3} [\frac{x ^{4}}{4}]_{- 1}^{2}

Compute the boundaries:

\frac{15}{4}

= \frac{1}{3} \cdot \frac{15}{4}

Simplify

= \frac{5}{4}

\int x e^{- 9 x} d x

(- 2 x y^{2} + 1 y) d x + (- 2 x^{2} y + 1 x) d y = 0

(x^{2} + 9)^{- 3}

\frac{d}{d x} (x sec^{2} (π x))

x \to - 0 lim (2 x e^{2 x} + k x)

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