Ashley Bell

2021-12-31

Evaluate the following integrals.

$\int {e}^{3-4x}dx$

Bubich13

Beginner2022-01-01Added 36 answers

Step 1

Given integral:

$\int {e}^{3-4x}dx$

Substitute t=3-4x and differentiate it w.r.t "x"

$\frac{dt}{dx}=\frac{d}{dx}(3-4x)$

$\frac{dt}{dx}=-4$

$\frac{-dt}{4}=dx$

Therefore, the given integral becomes,

$\frac{-1}{4}\int {e}^{t}dt$

Step 2

We know that

$\int {e}^{x}dx={e}^{x}+c$

Where, "c" is integration constant.

$\Rightarrow \frac{-1}{4}{e}^{t}+c$

Substitute t=3-4x in above equation, we get

$\Rightarrow \frac{-1}{4}{e}^{(3-4x)}+c$

$\Rightarrow \frac{-{e}^{(3-4x)}}{4}+c$

Given integral:

Substitute t=3-4x and differentiate it w.r.t "x"

Therefore, the given integral becomes,

Step 2

We know that

Where, "c" is integration constant.

Substitute t=3-4x in above equation, we get

Bubich13

Beginner2022-01-02Added 36 answers

Given

$\int {e}^{3-4x}dx$

$\int -\frac{{e}^{t}}{4}dt$

$-\frac{1}{4}\cdot \int {e}^{t}dt$

Solve

$-\frac{1}{4}{e}^{t}$

$-\frac{1}{4}{e}^{3-4x}$

$-\frac{{e}^{3-4x}}{4}$

Answer:

$-\frac{{e}^{3-4x}}{4}+C$

Solve

Answer:

Vasquez

Expert2022-01-07Added 669 answers

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