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

The integral from 0 to 1 of (x^2+x)e^x is e-1

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

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

e - 1

Decimal Notation

1.71828 \dots

Solution steps

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

Expand

(x^{2} + x) e^{x} : e^{x} x^{2} + e^{x} x

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int_{0}^{1} e^{x} x d x = 1

= e - 2 + 1

Simplify

= e - 1

\int_{0}^{1} e^{- t^{2}} d t

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 8 y = 0

sec^{2} (2 x + 5) \cdot 2

n \to 2 lim (n^{2})

f (x) = (5 x - 5) (2 x^{3} - x^{2} + 1)

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