What is the integral from 0 to 1 of e^x-e^{7x-6} ?

The integral from 0 to 1 of e^x-e^{7x-6} is e-1-(e^7-1)/(7e^6)

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integral from 0 to 1 of e^x-e^{7x-6}

\int_{0}^{1} e^{x} - e^{7 x - 6} d x

e - 1 - \frac{e ^{7} - 1}{7 e ^{6}}

Decimal

1.33030 \dots

Solution steps

\int_{0}^{1} e^{x} - e^{7 x - 6} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} e^{x} d x - \int_{0}^{1} e^{7 x - 6} d x

\int_{0}^{1} e^{x} d x = e - 1

\int_{0}^{1} e^{7 x - 6} d x = \frac{e ^{7} - 1}{7 e ^{6}}

= e - 1 - \frac{e ^{7} - 1}{7 e ^{6}}

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What is the integral from 0 to 1 of e^x-e^{7x-6} ?
The integral from 0 to 1 of e^x-e^{7x-6} is e-1-(e^7-1)/(7e^6)