Question

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

Answer 1

The answer is `=0.66`

Explanation:

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`

Answer 2

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

Explanation:

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!