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 Notation

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

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

= 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)

\int \frac{x}{2 x + 3} d x

\int x^{3} e^{9 x} d x

y = 10 x^{4} - 10 x^{2}, y = 7 x^{2}

x + y \cdot \frac{d y}{d x} = 0

(0.3) (4.1)

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)