Calculate $\int {5}^{x+1}{e}^{2x-1}\phantom{\rule{thinmathspace}{0ex}}dx$

klastiesym
2022-10-16
Answered

Calculate $\int {5}^{x+1}{e}^{2x-1}\phantom{\rule{thinmathspace}{0ex}}dx$

You can still ask an expert for help

getrdone07tl

Answered 2022-10-17
Author has **23** answers

Use integration by parts. Set $2x=y$ and you'll get:

$$5{e}^{-1}\int {5}^{\frac{y}{2}}{e}^{y}dy=5{e}^{-1}\int {5}^{\frac{y}{2}}d({e}^{y})$$

Then

$$5{e}^{-1}\int {5}^{\frac{y}{2}}{e}^{y}dy=5{e}^{-1}\int {5}^{\frac{y}{2}}d({e}^{y})=5{e}^{-1}({5}^{\frac{y}{2}}{e}^{y}-\frac{\mathrm{ln}(5)}{2}\int {5}^{\frac{y}{2}}{e}^{y}dy)$$

Combining the first and the last parts:

$$5{e}^{-1}\int {5}^{\frac{y}{2}}{e}^{y}dy=5{e}^{-1}({5}^{\frac{y}{2}}{e}^{y}-\frac{\mathrm{ln}(5)}{2}\int {5}^{\frac{y}{2}}{e}^{y}dy)$$

Cancelling out multipliers and collecting the integral terms:

$$(1+\frac{\mathrm{ln}(5)}{2})\int {5}^{\frac{y}{2}}{e}^{y}dy={5}^{\frac{y}{2}}{e}^{y}$$

So one shoulf get:

$$\int {5}^{\frac{y}{2}}{e}^{y}dy=\frac{{5}^{\frac{y}{2}}{e}^{y}}{1+\frac{\mathrm{ln}(5)}{2}}$$

So after that just remember the multiplier $5{e}^{-1}$ and the original variable $2x=y$

$$5{e}^{-1}\int {5}^{\frac{y}{2}}{e}^{y}dy=5{e}^{-1}\int {5}^{\frac{y}{2}}d({e}^{y})$$

Then

$$5{e}^{-1}\int {5}^{\frac{y}{2}}{e}^{y}dy=5{e}^{-1}\int {5}^{\frac{y}{2}}d({e}^{y})=5{e}^{-1}({5}^{\frac{y}{2}}{e}^{y}-\frac{\mathrm{ln}(5)}{2}\int {5}^{\frac{y}{2}}{e}^{y}dy)$$

Combining the first and the last parts:

$$5{e}^{-1}\int {5}^{\frac{y}{2}}{e}^{y}dy=5{e}^{-1}({5}^{\frac{y}{2}}{e}^{y}-\frac{\mathrm{ln}(5)}{2}\int {5}^{\frac{y}{2}}{e}^{y}dy)$$

Cancelling out multipliers and collecting the integral terms:

$$(1+\frac{\mathrm{ln}(5)}{2})\int {5}^{\frac{y}{2}}{e}^{y}dy={5}^{\frac{y}{2}}{e}^{y}$$

So one shoulf get:

$$\int {5}^{\frac{y}{2}}{e}^{y}dy=\frac{{5}^{\frac{y}{2}}{e}^{y}}{1+\frac{\mathrm{ln}(5)}{2}}$$

So after that just remember the multiplier $5{e}^{-1}$ and the original variable $2x=y$

Kasey Reese

Answered 2022-10-18
Author has **3** answers

Another way to solve:

$$\int {5}^{x+1}{e}^{2x-1}dx=\int \frac{5}{e}(5{e}^{2}{)}^{x}dx$$

$$(5{e}^{2}{)}^{x}=u\Rightarrow (5{e}^{2}{)}^{x}dx=\frac{du}{\mathrm{ln}(5{e}^{2})}$$

$$\int \frac{5}{e}(5{e}^{2}{)}^{x}dx=\frac{5}{e\mathrm{ln}(5{e}^{2})}\int du=\frac{5}{e\mathrm{ln}(5{e}^{2})}(u+C)$$

$$\int {5}^{x+1}{e}^{2x-1}dx=\int \frac{5}{e}(5{e}^{2}{)}^{x}dx$$

$$(5{e}^{2}{)}^{x}=u\Rightarrow (5{e}^{2}{)}^{x}dx=\frac{du}{\mathrm{ln}(5{e}^{2})}$$

$$\int \frac{5}{e}(5{e}^{2}{)}^{x}dx=\frac{5}{e\mathrm{ln}(5{e}^{2})}\int du=\frac{5}{e\mathrm{ln}(5{e}^{2})}(u+C)$$

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