What is the integral from 0 to 4 of (3^x)/(e^x) ?

The integral from 0 to 4 of (3^x)/(e^x) is (81)/(e^4(ln(3)-1))-1/(ln(3)-1)

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integral from 0 to 4 of (3^x)/(e^x)

\int_{0}^{4} \frac{3 ^{x}}{e ^{x}} d x

\frac{81}{e ^{4} ( ln ( 3 ) - 1 )} - \frac{1}{ln ( 3 ) - 1}

Decimal

4.90371 \dots

Solution steps

\int_{0}^{4} \frac{3 ^{x}}{e ^{x}} d x

Apply exponent rule:

\frac{a ^{c}}{b ^{c}} = (\frac{a}{b})^{c}

= \int_{0}^{4} (\frac{3}{e})^{x} d x

\int a^{x} d x = \frac{a ^{x}}{ln a}

= [\frac{( \frac{3}{e} ) ^{x}}{ln ( \frac{3}{e} )}]_{0}^{4}

Simplify

[\frac{( \frac{3}{e} ) ^{x}}{ln ( \frac{3}{e} )}]_{0}^{4} : [\frac{3 ^{x}}{( ln ( 3 ) - 1 ) e ^{x}}]_{0}^{4}

= [\frac{3 ^{x}}{( ln ( 3 ) - 1 ) e ^{x}}]_{0}^{4}

Compute the boundaries:

\frac{81}{e ^{4} ( ln ( 3 ) - 1 )} - \frac{1}{ln ( 3 ) - 1}

= \frac{81}{e ^{4} ( ln ( 3 ) - 1 )} - \frac{1}{ln ( 3 ) - 1}

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What is the integral from 0 to 4 of (3^x)/(e^x) ?
The integral from 0 to 4 of (3^x)/(e^x) is (81)/(e^4(ln(3)-1))-1/(ln(3)-1)