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

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

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

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

1

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

3

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

Simplify

= 1

\frac{d}{d x} (\frac{4 sin ( x )}{5 cos ( y )})

f (x) = 2 x^{2} - 5 x + 2, at x = a

x^{4} + 5 y^{4} \frac{x ^{5} + 4 d y}{d x} = 0, y (0) = 1

\frac{d y}{d x} - 2 x y = 0

\frac{( x - 10 )}{5 x + 7}

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