What is the integral from 1 to-1 of (3e^x) ?

The integral from 1 to-1 of (3e^x) is -3(e-1/e)

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integral from 1 to-1 of (3e^x)

\int_{1}^{- 1} (3 e^{x}) d x

- 3 (e - \frac{1}{e})

Decimal

- 7.05120 \dots

Solution steps

\int_{1}^{- 1} 3 e^{x} d x

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

e - \frac{1}{e}

= - 3 (e - \frac{1}{e})

\frac{d}{d x} (x e^{x^{3}})

\frac{1}{( s + 3 ) ^{2} + 4}

x^{2} y^{^{'}} - 2 x^{3} e^{- \frac{1}{x}} = y

\int \frac{sin ( x )}{cos ( 2 x )} d x

e^{t^{2}} y^{^{'}} + (t + t y^{2}) = 0

What is the integral from 1 to-1 of (3e^x) ?
The integral from 1 to-1 of (3e^x) is -3(e-1/e)