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

The integral from 0 to 1 of-1(x^2+1)e^{-x} is 6/e-3

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

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

\frac{6}{e} - 3

Decimal Notation

- 0.79272 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

: [- e^{- x} (x^{2} + 1) - \int - 2 e^{- x} x d x]_{0}^{1}

= - [- e^{- x} (x^{2} + 1) - \int - 2 e^{- x} x d x]_{0}^{1}

\int - 2 e^{- x} x d x = - 2 (- e^{- x} x - e^{- x})

= - [- e^{- x} (x^{2} + 1) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1}

Simplify

- [- e^{- x} (x^{2} + 1) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1} : - [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{1}

= - [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{6}{e} + 3

= - (- \frac{6}{e} + 3)

Simplify

= \frac{6}{e} - 3

\frac{\partial}{\partial x} (7 - 3 x^{2} - 3 y^{2})

(2.4) (6.12)

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

6 x + 5 cos (x)

x \to \infty lim (x^{2} + x)

What is the integral from 0 to 1 of-1(x^2+1)e^{-x} ?
The integral from 0 to 1 of-1(x^2+1)e^{-x} is 6/e-3