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

Question

Emilia Aguilar · Accepted Answer

The answer is #=0.66#

Let #y=5^x#

Taking log on both sides

#lny=xln5#

#y=e^(xln5)#

Therefore,

#int5^xdx=inte^(xln5)dx=e^(xln5)/ln5=5^x/ln5#

Similarly,

#y=3^x#

Taking log on both sides

#lny=xln3#

#y=e^(xln3)#

Therefore,

#int3^xdx=inte^(xln3)dx=e^(xln3)/ln3=3^x/ln3#

Therefore,

#int_0^1(5^x-3^x)dx=int_0^1 5^xdx-int_0^1 3^xdx#

#=[5^x/ln5-3^x/ln3]_0^1#

#=(5/ln5-3/ln3)-(1/ln5-1/ln3)#

#=4/ln5-2/ln3#

#=0.66#

Elijah Abram · Answer

The integral has value #4/ln(5) - 2/ln(3)#

Separating the integrals, we get:

#int_0^1 5^xdx - int_0^1 3^xdx#

Now use the formula #int(a^x)dx = a^x/ln(a)#.

#[5^x/ln5]_0^1 - [3^x/ln3]_0^1#

#5/ln(5) - 5^0/ln(5) - (3/ln3 - 3^0/ln3)#

#4/ln(5) - 2/ln(3)#

Hopefully this helps!

Aurora Alvarado · Answer

undefined To integrate $ \int_{0}^{1} 5^x - 3^x \, dx $, you can use the properties of exponents to simplify the integral.

First, split the integral into two separate integrals:

$$ \int_{0}^{1} 5^x \, dx - \int_{0}^{1} 3^x \, dx $$

Now, integrate each term separately:

For $ \int_{0}^{1} 5^x \, dx $:
$$ \int_{0}^{1} 5^x \, dx = \left[ \frac{5^x}{\ln(5)} \right]_{0}^{1} = \frac{5^1}{\ln(5)} - \frac{5^0}{\ln(5)} = \frac{5}{\ln(5)} - \frac{1}{\ln(5)} = \frac{4}{\ln(5)} $$

For $ \int_{0}^{1} 3^x \, dx $:
$$ \int_{0}^{1} 3^x \, dx = \left[ \frac{3^x}{\ln(3)} \right]_{0}^{1} = \frac{3^1}{\ln(3)} - \frac{3^0}{\ln(3)} = \frac{3}{\ln(3)} - \frac{1}{\ln(3)} = \frac{2}{\ln(3)} $$

Substitute these results back into the original expression:

$$ \int_{0}^{1} 5^x - 3^x \, dx = \frac{4}{\ln(5)} - \frac{2}{\ln(3)} $$