# How do you integrate int 5^x-3^xdx from [0,1]?

May 5, 2017

#### Answer:

The answer is $= 0.66$

#### Explanation:

Let $y = {5}^{x}$

Taking log on both sides

$\ln y = x \ln 5$

$y = {e}^{x \ln 5}$

Therefore,

$\int {5}^{x} \mathrm{dx} = \int {e}^{x \ln 5} \mathrm{dx} = {e}^{x \ln 5} / \ln 5 = {5}^{x} / \ln 5$

Similarly,

$y = {3}^{x}$

Taking log on both sides

$\ln y = x \ln 3$

$y = {e}^{x \ln 3}$

Therefore,

$\int {3}^{x} \mathrm{dx} = \int {e}^{x \ln 3} \mathrm{dx} = {e}^{x \ln 3} / \ln 3 = {3}^{x} / \ln 3$

Therefore,

${\int}_{0}^{1} \left({5}^{x} - {3}^{x}\right) \mathrm{dx} = {\int}_{0}^{1} {5}^{x} \mathrm{dx} - {\int}_{0}^{1} {3}^{x} \mathrm{dx}$

$= {\left[{5}^{x} / \ln 5 - {3}^{x} / \ln 3\right]}_{0}^{1}$

$= \left(\frac{5}{\ln} 5 - \frac{3}{\ln} 3\right) - \left(\frac{1}{\ln} 5 - \frac{1}{\ln} 3\right)$

$= \frac{4}{\ln} 5 - \frac{2}{\ln} 3$

$= 0.66$

May 5, 2017

#### Answer:

The integral has value $\frac{4}{\ln} \left(5\right) - \frac{2}{\ln} \left(3\right)$

#### Explanation:

Separating the integrals, we get:

${\int}_{0}^{1} {5}^{x} \mathrm{dx} - {\int}_{0}^{1} {3}^{x} \mathrm{dx}$

Now use the formula $\int \left({a}^{x}\right) \mathrm{dx} = {a}^{x} / \ln \left(a\right)$.

${\left[{5}^{x} / \ln 5\right]}_{0}^{1} - {\left[{3}^{x} / \ln 3\right]}_{0}^{1}$

$\frac{5}{\ln} \left(5\right) - {5}^{0} / \ln \left(5\right) - \left(\frac{3}{\ln} 3 - {3}^{0} / \ln 3\right)$

$\frac{4}{\ln} \left(5\right) - \frac{2}{\ln} \left(3\right)$

Hopefully this helps!