What is the integral from 4 to 1 of 2e^{(x+3)} ?

The integral from 4 to 1 of 2e^{(x+3)} is -2e^7+2e^4

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integral from 4 to 1 of 2e^{(x+3)}

\int_{4}^{1} 2 e^{(x + 3)} d x

- 2 e^{7} + 2 e^{4}

Decimal Notation

- 2084.07001 \dots

Solution steps

\int_{4}^{1} 2 e^{x + 3} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{4} 2 e^{x + 3} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2 \cdot \int_{1}^{4} e^{x + 3} d x

e^{x + 3} = e^{3} e^{x}

= - 2 \cdot \int_{1}^{4} e^{3} e^{x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2 e^{3} \cdot \int_{1}^{4} e^{x} d x

Use the common integral:

\int e^{x} d x = e^{x}

= - 2 e^{3} [e^{x}]_{1}^{4}

Compute the boundaries:

e^{4} - e

= - 2 e^{3} (e^{4} - e)

Simplify

= - 2 e^{7} + 2 e^{4}

\frac{d}{d x} (ln (4) (x))

y^{^{''}} - 2 y^{^{'}} = 8 sin (x) + 2 x e^{x}

\int (x + x^{2}) d x

\frac{d}{d x} (4 - 5 sin (2 x + \frac{π}{4}))

f (x) = 2 x 1 - 3 x, at x = - 2

What is the integral from 4 to 1 of 2e^{(x+3)} ?
The integral from 4 to 1 of 2e^{(x+3)} is -2e^7+2e^4