What is the integral from 0 to 1 of (5x^e+4e^x) ?

The integral from 0 to 1 of (5x^e+4e^x) is 5/(e+1)+4(e-1)

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integral from 0 to 1 of (5x^e+4e^x)

\int_{0}^{1} (5 x^{e} + 4 e^{x}) d x

\frac{5}{e + 1} + 4 (e - 1)

Decimal Notation

8.21783 \dots

Solution steps

\int_{0}^{1} 5 x^{e} + 4 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} 5 x^{e} d x + \int_{0}^{1} 4 e^{x} d x

\int_{0}^{1} 5 x^{e} d x = \frac{5}{e + 1}

\int_{0}^{1} 4 e^{x} d x = 4 (e - 1)

= \frac{5}{e + 1} + 4 (e - 1)

\int_{0}^{3} (x^{2} - 5 x + 9) d x

\int_{2}^{4} x^{2} + 5 x + 7 + 1 d x

\int_{- 7}^{4} \frac{1}{2} x d x

\int_{3}^{6} (2 x + 9) d x

\int_{2}^{7} 3 x - 1 d x

What is the integral from 0 to 1 of (5x^e+4e^x) ?
The integral from 0 to 1 of (5x^e+4e^x) is 5/(e+1)+4(e-1)