What is the integral from 0 to 1 of 4xy ?

The integral from 0 to 1 of 4xy is 2x

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integral from 0 to 1 of 4xy

\int_{0}^{1} 4 x y d y

2 x

Solution steps

\int_{0}^{1} 4 x y d y

Take the constant out:

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

= 4 x \cdot \int_{0}^{1} y d y

Apply the Power Rule

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

= 4 x [\frac{y ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= 4 x \frac{1}{2}

Simplify

= 2 x

\int \frac{sin ( 6 e ^{- 5 x} )}{e ^{5 x}} d x

h \to 0 lim (\frac{( 2 + h ) ^{2} - 4}{h})

n = 1 \sum \infty n^{4} (1 + n^{5})^{2}

x \to 0 lim (\frac{sin ( 40 )}{x})

f (x) = \frac{3}{x + 3}