What is the integral from 0 to 1 of (7x^e+2e^x) ?

The integral from 0 to 1 of (7x^e+2e^x) is 7/(e+1)+2(e-1)

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integral from 0 to 1 of (7x^e+2e^x)

\int_{0}^{1} (7 x^{e} + 2 e^{x}) d x

\frac{7}{e + 1} + 2 (e - 1)

Decimal Notation

5.31915 \dots

Solution steps

\int_{0}^{1} 7 x^{e} + 2 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} 7 x^{e} d x + \int_{0}^{1} 2 e^{x} d x

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

\int_{0}^{1} 2 e^{x} d x = 2 (e - 1)

= \frac{7}{e + 1} + 2 (e - 1)

\frac{d y}{d x} = x x - 27

9 x - 6 x, at x = 1

\int csc^{4} (x) cot^{4} (x) d x

x \to 0 lim ((2 x + 1)^{\frac{1}{x}})

\frac{d}{d x} (x^{3} - 6 x^{2} - 10 x)

What is the integral from 0 to 1 of (7x^e+2e^x) ?
The integral from 0 to 1 of (7x^e+2e^x) is 7/(e+1)+2(e-1)