What is the integral from 0 to 1 of (9x^e+2e^x) ?

The integral from 0 to 1 of (9x^e+2e^x) is 9/(e+1)+2(e-1)

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integral from 0 to 1 of (9x^e+2e^x)

\int_{0}^{1} (9 x^{e} + 2 e^{x}) d x

\frac{9}{e + 1} + 2 (e - 1)

Decimal Notation

5.85703 \dots

Solution steps

\int_{0}^{1} 9 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} 9 x^{e} d x + \int_{0}^{1} 2 e^{x} d x

\int_{0}^{1} 9 x^{e} d x = \frac{9}{e + 1}

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

= \frac{9}{e + 1} + 2 (e - 1)

\frac{d y}{d x} = 2 x y^{2} + 3 x^{2} y^{2}

f (t) = \frac{cos ( t )}{t ^{7}}

x \to 2 lim (5 x - 2)

\frac{d}{d x} (x cos (y - x))

\int cos (4 x + 5) d x

What is the integral from 0 to 1 of (9x^e+2e^x) ?
The integral from 0 to 1 of (9x^e+2e^x) is 9/(e+1)+2(e-1)