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

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

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

\int_{0}^{1} 6 (x^{2} + 1) e^{- x} d x

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

Decimal Notation

4.75634 \dots

Solution steps

\int_{0}^{1} 6 (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

= 6 \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}

= 6 [- 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})

= 6 [- e^{- x} (x^{2} + 1) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1}

Simplify

6 [- e^{- x} (x^{2} + 1) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1} : 6 [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{1}

= 6 [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{6}{e} + 3

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

\frac{d y}{d t} + 20 y = 24

\int 4 θ sec^{2} (θ) d θ

y^{^{'}} = - \frac{y ^{2}}{( 0.05 + y )}, y (0) = 9

\int \frac{1}{2} x^{\frac{3}{4}} d x

\int (9) d x

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