What is the integral from 0 to 1 of (x^e+e^x) ?

The integral from 0 to 1 of (x^e+e^x) is 1/(e+1)+e-1

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integral from 0 to 1 of (x^e+e^x)

\int_{0}^{1} (x^{e} + e^{x}) d x

\frac{1}{e + 1} + e - 1

Solution steps

\int_{0}^{1} x^{e} + e^{x} 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} x^{e} d x + \int_{0}^{1} e^{x} d x

\int_{0}^{1} x^{e} d x = \frac{1}{e + 1}

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

= \frac{1}{e + 1} + e - 1

x^{10} h (x)

f (x) = sec^{2} (x) - 1

\int - 16 x^{2} d x

\int t^{2} sin (t^{3}) d t

\int (x - 1)^{10} (x + 3)^{2} d x

What is the integral from 0 to 1 of (x^e+e^x) ?
The integral from 0 to 1 of (x^e+e^x) is 1/(e+1)+e-1