What is the integral from x to 5x of e^{3t} ?

The integral from x to 5x of e^{3t} is 1/3 (e^{15x}-e^{3x})

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integral from x to 5x of e^{3t}

\int_{x}^{5 x} e^{3 t} d t

\frac{1}{3} (e^{15 x} - e^{3 x})

Solution steps

\int_{x}^{5 x} e^{3 t} d t

Apply u-substitution

: \int_{3 x}^{15 x} e^{u} \frac{1}{3} d u

= \int_{3 x}^{15 x} 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

= \frac{1}{3} \cdot \int_{3 x}^{15 x} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{3 x}^{15 x}

Compute the boundaries:

e^{15 x} - e^{3 x}

= \frac{1}{3} (e^{15 x} - e^{3 x})

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What is the integral from x to 5x of e^{3t} ?
The integral from x to 5x of e^{3t} is 1/3 (e^{15x}-e^{3x})