What is the integral from ln(4) to ln(9) of e^{x/2} ?

The integral from ln(4) to ln(9) of e^{x/2} is 2

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integral from ln(4) to ln(9) of e^{x/2}

\int_{l n (4)}^{l n (9)} e^{\frac{x}{2}} d x

2

Solution steps

\int_{l n (4)}^{l n (9)} e^{\frac{x}{2}} d x

Apply u-substitution

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

= \int_{l n (2)}^{l n (3)} 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 (2)}^{l n (3)} e^{u} d u

Use the common integral:

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

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

Compute the boundaries:

1

= 2 \cdot 1

Simplify

= 2

f (x) = sin (\frac{1}{x ^{2}})

\int \frac{cos ^{6} ( x )}{sin ^{10} ( x )} d x

y^{^{''}} + \frac{a _{1^{2}} π ^{2}}{a _{2}^{2}} a_{3}^{2} y = 0

sin (s)

\int - e^{x} + sin (2 x) d x

What is the integral from ln(4) to ln(9) of e^{x/2} ?
The integral from ln(4) to ln(9) of e^{x/2} is 2