## ksaimouli 2 years ago integral

1. ksaimouli

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2. ksaimouli

@JamesJ

3. ksaimouli

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4. robtobey

$\text{The} \text{ derivative} \text{ of } 5^{x } \text { is}:$$5^x\text{ Log}[5]$

5. ksaimouli

not sure i can do this because (5^x) (ln5) separate

6. amoodarya

use u=5^x

7. ksaimouli

then integral of u is (5^x)/ln5 right

8. ksaimouli

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9. ksaimouli

if i use u substitution ln5 goes to top so no use

10. Spacelimbus

$\Large \int 5^x\ln 5 dx$ ?? yes `

11. ksaimouli

yes

12. Spacelimbus

$\Large \ln 5 \int 5^xdx = \ln 5 \int e^{\ln5x}dx$

13. Spacelimbus

If I understand the integral correctly.

14. Spacelimbus

Just rewrite the number 5 as e^ln5 and then you have a regular exponential function (linear) you can substitute and evaluate.

15. ksaimouli

can u really do that manuplation

16. Spacelimbus

Of course, here's the proof: $\Large a^b=c \\ \\ \Large b\log_aa= \log_ac$that leads to : $\Large b=\log_ac$ And back substitution gives: $\Large a^{\log_ac}=c$ Just a different way of writing numbers, that's also the reason why a lot of people write the exponential decay with as an e function, rather then the regular y(t)=ba^t form

17. Spacelimbus

Integration result gives you $\Large e^{\ln5 x}+C$ Deriving with respect to x: $\Large \ln5 \cdot e^{\ln 5 x} = \ln 5 \cdot \left(e^{\ln5} \right)^x = \ln5 \cdot 5^x$