What is the integral from ln(16) to ln(49) of e^{x/2} ?

The integral from ln(16) to ln(49) of e^{x/2} is 6

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integral from ln(16) to ln(49) of e^{x/2}

\int_{l n (16)}^{l n (49)} e^{\frac{x}{2}} d x

6

Solution steps

\int_{l n (16)}^{l n (49)} e^{\frac{x}{2}} d x

Apply u-substitution

: \int_{2 l n (2)}^{l n (7)} e^{u} \cdot 2 d u

= \int_{2 l n (2)}^{l n (7)} e^{u} \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int_{2 l n (2)}^{l n (7)} e^{u} d u

Use the common integral:

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

= 2 [e^{u}]_{2 l n (2)}^{l n (7)}

Compute the boundaries:

3

= 2 \cdot 3

Simplify

= 6

\int \frac{1}{x ^{2} + 2 x + 17} d x

\frac{1}{x - 1}, 16

\frac{d}{d x} (∣ 0 ∣)

\frac{d}{d x} (1 + 6 x)

\frac{d}{d x} (∣ 2 x ∣)

What is the integral from ln(16) to ln(49) of e^{x/2} ?
The integral from ln(16) to ln(49) of e^{x/2} is 6