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

The integral from 0 to 1 of (x^2+1)e^{-1} is 4/(3e)

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integral from 0 to 1 of (x^2+1)e^{-1}

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

\frac{4}{3 e}

Decimal Notation

0.49050 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= e^{- 1} \cdot \int_{0}^{1} x^{2} + 1 d x

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{e} \cdot \int_{0}^{1} x^{2} + 1 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

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

\int_{0}^{1} x^{2} d x = \frac{1}{3}

\int_{0}^{1} 1 d x = 1

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

Simplify

\frac{1}{e} (\frac{1}{3} + 1) : \frac{4}{3 e}

= \frac{4}{3 e}

\int \frac{5}{1000 + t} d t

\int x^{2} (4 - x) d x

x \to 2 lim (\frac{4 - x ^{2}}{2 + x})

\frac{d}{d x} (20 x - x^{2})

\frac{d x}{d t} = \frac{3}{2} t cos^{2} (x), x (0) = - 1.4

What is the integral from 0 to 1 of (x^2+1)e^{-1} ?
The integral from 0 to 1 of (x^2+1)e^{-1} is 4/(3e)