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

Decimal Notation

1.98722 \dots

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

y = x 5^{x + 4}

\int_{0}^{2 π} e^{i t} d t

x \to 0 lim (\frac{x _{- 1}}{x})

6 sin^{2} (t)

f (x) = sin (x) tan (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