What is the integral from 0 to 1 of 40x ?

The integral from 0 to 1 of 40x is 20

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integral from 0 to 1 of 40x

\int_{0}^{1} 40 x d x

20

Solution steps

\int_{0}^{1} 40 x d x

Take the constant out:

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

= 40 \cdot \int_{0}^{1} x d x

Apply the Power Rule

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

= 40 [\frac{x ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= 40 \cdot \frac{1}{2}

Simplify

= 20

x \to 0 lim (+ (6 - \frac{1}{x ^{3}}))

n = 3 \sum \infty (\frac{8}{11})^{- n}

\int (t^{2} + 4 t) d t

n = 1 \sum \infty \frac{1}{2 ^{n}}

\frac{\partial}{\partial y} (n sin (2 (m x + n y)))