What is the integral from-1 to 2 of (x^2)/8 ?

The integral from-1 to 2 of (x^2)/8 is 3/8

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integral from-1 to 2 of (x^2)/8

\int_{- 1}^{2} \frac{x ^{2}}{8} d x

\frac{3}{8}

Decimal Notation

0.375

Solution steps

\int_{- 1}^{2} \frac{x ^{2}}{8} d x

Take the constant out:

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

= \frac{1}{8} \cdot \int_{- 1}^{2} x^{2} d x

Apply the Power Rule

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

= \frac{1}{8} [\frac{x ^{3}}{3}]_{- 1}^{2}

Compute the boundaries:

3

= \frac{1}{8} \cdot 3

Simplify

= \frac{3}{8}

\frac{e t ^{4}}{24}

\frac{d x}{d t} + x e^{t} = 1 - x

(5, - 6), (8, - 4)

x \to 0 lim (1 - 3 x)

x y^{^{'}} + \frac{1}{2} y = x

What is the integral from-1 to 2 of (x^2)/8 ?
The integral from-1 to 2 of (x^2)/8 is 3/8