What is the integral from 0 to 1 of 7(x^2+1)e^{-x} ?

The integral from 0 to 1 of 7(x^2+1)e^{-x} is 7(-6/e+3)

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integral from 0 to 1 of 7(x^2+1)e^{-x}

\int_{0}^{1} 7 (x^{2} + 1) e^{- x} d x

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

Decimal Notation

5.54906 \dots

Solution steps

\int_{0}^{1} 7 (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

= 7 \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}

= 7 [- 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})

= 7 [- e^{- x} (x^{2} + 1) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1}

Simplify

7 [- e^{- x} (x^{2} + 1) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1} : 7 [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{1}

= 7 [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{6}{e} + 3

= 7 (- \frac{6}{e} + 3)

\int \frac{x ^{3} - 8 x}{x ^{2}} d x

\frac{d}{d x} ((x^{2} + sin (x)) sec (x))

\frac{\partial}{\partial t} (s + t)

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 sin (2 t)

- 2 x^{2} y^{^{''}} + 3 x y^{^{'}} - y = 0

What is the integral from 0 to 1 of 7(x^2+1)e^{-x} ?
The integral from 0 to 1 of 7(x^2+1)e^{-x} is 7(-6/e+3)