What is the integral from 0 to 7 of 2e^x ?

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

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integral from 0 to 7 of 2e^x

\int_{0}^{7} 2 e^{x} d x

2 (e^{7} - 1)

Decimal Notation

2191.26631 \dots

Solution steps

\int_{0}^{7} 2 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}^{7} e^{x} d x

Use the common integral:

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

= 2 [e^{x}]_{0}^{7}

Compute the boundaries:

e^{7} - 1

= 2 (e^{7} - 1)

n = 2 \sum \infty \frac{1}{( n ^{3} - n )}

\frac{0.32}{( 15.5 \cdot s + 1 ) \cdot s}

\frac{1}{( s ^{2} + 167 s + 12874 )}

x \to 3 lim (x + 8)

\frac{d y}{d x} = \frac{1}{3} y (8 - y), y (0) = 1

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