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

The integral from 0 to 1 of-9(x^2+1)e^{-x} is -9(-6/e+3)

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

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

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

Decimal

- 7.13451 \dots

Solution steps

\int_{0}^{1} - 9 (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

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

Apply Integration By Parts

= - 9 [- 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})

= - 9 [- e^{- x} (x^{2} + 1) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1}

Simplify

- 9 [- e^{- x} (x^{2} + 1) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1} : - 9 [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{1}

= - 9 [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{6}{e} + 3

= - 9 (- \frac{6}{e} + 3)

t an g e n t y = (1 + 2 x)^{2}, (1, 9)

\frac{\partial}{\partial x} (arctan (y x))

\frac{d}{d x} (1 + x)

x y^{^{'}} - y = x^{2}

x \to 7 lim (x - 7)

What is the integral from 0 to 1 of-9(x^2+1)e^{-x} ?
The integral from 0 to 1 of-9(x^2+1)e^{-x} is -9(-6/e+3)