What is the integral from 1 to 2 of 4e^{3x-2} ?

The integral from 1 to 2 of 4e^{3x-2} is (4(e^4-e))/3

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integral from 1 to 2 of 4e^{3x-2}

\int_{1}^{2} 4 e^{3 x - 2} d x

\frac{4 ( e ^{4} - e )}{3}

Decimal

69.17315 \dots

Solution steps

\int_{1}^{2} 4 e^{3 x - 2} d x

Take the constant out:

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

= 4 \cdot \int_{1}^{2} e^{3 x - 2} d x

Apply u-substitution

: \int_{1}^{4} e^{u} \frac{1}{3} d u

= 4 \cdot \int_{1}^{4} e^{u} \frac{1}{3} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int_{1}^{4} e^{u} d u

Use the common integral:

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

= 4 \cdot \frac{1}{3} [e^{u}]_{1}^{4}

Simplify

4 \cdot \frac{1}{3} [e^{u}]_{1}^{4} : \frac{4}{3} [e^{u}]_{1}^{4}

= \frac{4}{3} [e^{u}]_{1}^{4}

Compute the boundaries:

e^{4} - e

= \frac{4}{3} (e^{4} - e)

Simplify

= \frac{4 ( e ^{4} - e )}{3}

\int_{3}^{10} 1

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\int_{1}^{4} \frac{ln ( x )}{x ^{3}} d x

\int_{0}^{l n (2)} 6 e^{2 t} d t

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

What is the integral from 1 to 2 of 4e^{3x-2} ?
The integral from 1 to 2 of 4e^{3x-2} is (4(e^4-e))/3