What is the integral from 0 to 1 of (8x^2-5)e^x ?

The integral from 0 to 1 of (8x^2-5)e^x is 3e-11

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integral from 0 to 1 of (8x^2-5)e^x

\int_{0}^{1} (8 x^{2} - 5) e^{x} d x

3 e - 11

Decimal Notation

- 2.84515 \dots

Solution steps

\int_{0}^{1} (8 x^{2} - 5) e^{x} d x

Apply Integration By Parts

: [(8 x^{2} - 5) e^{x} - \int 16 x e^{x} d x]_{0}^{1}

= [(8 x^{2} - 5) e^{x} - \int 16 x e^{x} d x]_{0}^{1}

\int 16 x e^{x} d x = 16 (e^{x} x - e^{x})

= [(8 x^{2} - 5) e^{x} - 16 (e^{x} x - e^{x})]_{0}^{1}

Expand

(8 x^{2} - 5) e^{x} - 16 (e^{x} x - e^{x}) : 8 e^{x} x^{2} - 16 e^{x} x + 11 e^{x}

= [8 e^{x} x^{2} - 16 e^{x} x + 11 e^{x}]_{0}^{1}

Compute the boundaries:

3 e - 11

= 3 e - 11

\frac{\partial}{\partial y} (2 y t + x j)

\int \frac{x ^{2}}{16 + x ^{2}} d x

\int 4 t^{3} y - 15 t^{2} - y d t

11 x - 3, 10 x - 5, [12, 34]

\int \frac{1}{t ^{2} + 1} d t

What is the integral from 0 to 1 of (8x^2-5)e^x ?
The integral from 0 to 1 of (8x^2-5)e^x is 3e-11