What is the integral from 0 to 1 of (3x^e+8e^x) ?

The integral from 0 to 1 of (3x^e+8e^x) is 3/(e+1)+8(e-1)

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integral from 0 to 1 of (3x^e+8e^x)

\int_{0}^{1} (3 x^{e} + 8 e^{x}) d x

\frac{3}{e + 1} + 8 (e - 1)

Decimal

14.55307 \dots

Solution steps

\int_{0}^{1} 3 x^{e} + 8 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} 3 x^{e} d x + \int_{0}^{1} 8 e^{x} d x

\int_{0}^{1} 3 x^{e} d x = \frac{3}{e + 1}

\int_{0}^{1} 8 e^{x} d x = 8 (e - 1)

= \frac{3}{e + 1} + 8 (e - 1)

\int_{0}^{2} 3 x (x - 4) d x

\int_{0}^{\infty} \frac{3}{x ^{2} + 5} d x

\int_{4}^{10} 7 d x

\int_{- 1}^{0} \frac{1}{x ^{2} + 1} d x

\int_{0}^{7} t^{3} - 9 t^{2} + 20 t d t

What is the integral from 0 to 1 of (3x^e+8e^x) ?
The integral from 0 to 1 of (3x^e+8e^x) is 3/(e+1)+8(e-1)