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

The integral from 0 to 1 of (8x^2-2)e^x is 2(3e-7)

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

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

2 (3 e - 7)

Decimal Notation

2.30969 \dots

Solution steps

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

(8 x^{2} - 2) e^{x} = 2 (4 x^{2} - 1) e^{x}

= \int_{0}^{1} 2 (4 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

= 2 \cdot \int_{0}^{1} (4 x^{2} - 1) e^{x} d x

Apply Integration By Parts

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

= 2 [(4 x^{2} - 1) e^{x} - \int 8 x e^{x} d x]_{0}^{1}

\int 8 x e^{x} d x = 8 (e^{x} x - e^{x})

= 2 [(4 x^{2} - 1) e^{x} - 8 (e^{x} x - e^{x})]_{0}^{1}

Expand

(4 x^{2} - 1) e^{x} - 8 (e^{x} x - e^{x}) : 4 e^{x} x^{2} - 8 e^{x} x + 7 e^{x}

= 2 [4 e^{x} x^{2} - 8 e^{x} x + 7 e^{x}]_{0}^{1}

Compute the boundaries:

3 e - 7

= 2 (3 e - 7)

\frac{d}{d t} ((arccos (t))^{2})

\int 2 cos^{3} (4 x) d x

\frac{\partial}{\partial v} (2 uv)

\frac{f ( x )}{x ^{7}}

x \to 0 lim ((5 + x)^{x})

What is the integral from 0 to 1 of (8x^2-2)e^x ?
The integral from 0 to 1 of (8x^2-2)e^x is 2(3e-7)