What is the integral from 0 to 1 of 6e^{2x} ?

The integral from 0 to 1 of 6e^{2x} is 3(e^2-1)

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integral from 0 to 1 of 6e^{2x}

\int_{0}^{1} 6 e^{2 x} d x

3 (e^{2} - 1)

Solution steps

\int_{0}^{1} 6 e^{2 x} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} e^{2 x} d x

Apply u-substitution

: \int_{0}^{2} e^{u} \frac{1}{2} d u

= 6 \cdot \int_{0}^{2} e^{u} \frac{1}{2} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int_{0}^{2} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= 6 \cdot \frac{1}{2} [e^{u}]_{0}^{2}

Simplify

6 \cdot \frac{1}{2} [e^{u}]_{0}^{2} : 3 [e^{u}]_{0}^{2}

= 3 [e^{u}]_{0}^{2}

Compute the boundaries:

e^{2} - 1

= 3 (e^{2} - 1)

\int \frac{1}{x ( 1 + 4 x ^{2} )} d x

x \to + 0 - lim (arctan (\frac{1}{x}))

\frac{- a}{x ^{2}} + \frac{a}{( x - 2 ) ^{2}}

f (x) = \frac{2 x + 3}{x ^{2} + 3 x + 15}

\int_{0}^{2} \frac{1}{4 - 2 x} d x

What is the integral from 0 to 1 of 6e^{2x} ?
The integral from 0 to 1 of 6e^{2x} is 3(e^2-1)