What is the integral from ln(9) to ln(36) of e^{x/2} ?

The integral from ln(9) to ln(36) of e^{x/2} is 6

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integral from ln(9) to ln(36) of e^{x/2}

\int_{l n (9)}^{l n (36)} e^{\frac{x}{2}} d x

6

Solution steps

\int_{l n (9)}^{l n (36)} e^{\frac{x}{2}} d x

Apply u-substitution

: \int_{l n (3)}^{l n (6)} e^{u} \cdot 2 d u

= \int_{l n (3)}^{l n (6)} 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_{l n (3)}^{l n (6)} e^{u} d u

Use the common integral:

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

= 2 [e^{u}]_{l n (3)}^{l n (6)}

Compute the boundaries:

3

= 2 \cdot 3

Simplify

= 6

f (x) = \frac{3}{x ^{4}}

\int e^{x^{2} - 4} d x

x \to 4 lim (3 - 4)

\int \frac{1}{1 + 16 x} d x

\frac{d y}{d x} - \frac{y}{x} = 7 x e^{x}

What is the integral from ln(9) to ln(36) of e^{x/2} ?
The integral from ln(9) to ln(36) of e^{x/2} is 6