What is the integral of 0.4^{x/2} ?

The integral of 0.4^{x/2} is -2.18271…*0.4^{x/2}+C

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integral of 0.4^{x/2}

\int 0. 4^{\frac{x}{2}} d x

- 2.18271 \dots \cdot 0. 4^{\frac{x}{2}} + C

Solution steps

\int 0. 4^{\frac{x}{2}} d x

Apply u-substitution

: \int 0. 4^{u} \cdot 2 d u

= \int 0. 4^{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 0. 4^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= 2 \cdot \frac{0. 4 ^{u}}{ln ( 0.4 )}

Substitute back

u = \frac{x}{2}

= 2 \cdot \frac{0. 4 ^{\frac{x}{2}}}{ln ( 0.4 )}

Simplify

2 \cdot \frac{0. 4 ^{\frac{x}{2}}}{ln ( 0.4 )} : - 2.18271 \dots \cdot 0. 4^{\frac{x}{2}}

= - 2.18271 \dots \cdot 0. 4^{\frac{x}{2}}

Add a constant to the solution

= - 2.18271 \dots \cdot 0. 4^{\frac{x}{2}} + C

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What is the integral of 0.4^{x/2} ?
The integral of 0.4^{x/2} is -2.18271…*0.4^{x/2}+C