What is the integral from ln(4) to ln(16) of e^{x/2} ?

The integral from ln(4) to ln(16) of e^{x/2} is 4

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integral from ln(4) to ln(16) of e^{x/2}

\int_{l n (4)}^{l n (16)} e^{\frac{x}{2}} d x

4

Solution steps

\int_{l n (4)}^{l n (16)} e^{\frac{x}{2}} d x

Apply u-substitution

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

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

Use the common integral:

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

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

Compute the boundaries:

2

= 2 \cdot 2

Simplify

= 4

3 x e^{- 2 x}

G (r) = r + 7 r

\frac{d ^{5}}{d x ^{5}} (sin (x))

x \cdot x

\int 5 + 8 x d x

What is the integral from ln(4) to ln(16) of e^{x/2} ?
The integral from ln(4) to ln(16) of e^{x/2} is 4