What is the integral from 0 to 1 of 9x ?

The integral from 0 to 1 of 9x is 9/2

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integral from 0 to 1 of 9x

\int_{0}^{1} 9 x d x

\frac{9}{2}

Decimal Notation

4.5

Solution steps

\int_{0}^{1} 9 x d x

Take the constant out:

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

= 9 \cdot \int_{0}^{1} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{1}

= 9 [\frac{x ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= 9 \cdot \frac{1}{2}

Simplify

= \frac{9}{2}

\int x arcsec (x) d x

\int \frac{( x e ^{2 x} )}{( 1 + 2 x ) ^{2}} d x

f (x) = \frac{1}{x}, at x = - 2

\frac{d}{d x} (\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 6 x)

\frac{d}{d x} (2 e^{x^{2}})