The answer is : int_0^1 x5^x dx ~~1.56∫10x5xdx≈1.56.
We have h(x) = 5^x x = f'(x)g(x).
Here, f'(x) = 5^x and g(x) = x.
The antiderivative of h(x) is given by :
f(x)g(x) - int f(x)g'(x)dx.
Firstly, let's calculate the derivative of 5^x by using the limit definition, it will help us to get its antiderivative :
j'(x) = lim_(h->0)(j(x + h) -j(x))/h
(5^x)' = lim_(h->0) (5^(x+h) -5^(x))/h = lim_(h->0) (5^(x)*(5^h-1))/h
= 5^(x) lim_(h->0) (5^h-1)/h = 5^(x) lim_(h->0) (e^(hln(5))-1)/h
= 5^(x) lim_(h->0) ln(5)((e^(hln(5))-1)/(hln(5)))
= 5^(x) ln(5) lim_(h->0) (e^(hln(5))-1)/(hln(5))
=5^(x) ln(5) lim_(k->0) (e^(k)-1)/(k) , where k = hln(5)
By the definition of e, lim_(k->0) (e^(k)-1)/(k)=1.
So (5^x)' = 5^x ln(5).
Therefore, the antiderivative of f'(x) = 5^x is f(x) = 5^x/ln(5) because :
(5^x/ln(5))' = ((5^x)')/ln(5) = (5^x ln(5))/ln(5) = 5^x
And the derivative of g(x) = x is g'(x) = 1.
Thus, the antiderivative of h(x) is :
H(x) = f(x)g(x) - int f(x)g'(x)dx = (5^x x)/ln(5) - int 5^x/ln(5)
= (5^x x)/ln(5) - 1/ln(5) int 5^x = (5^x x)/ln(5) - 5^x/ln^2(5)
Now, we can calculate the integral :
int_0^1 h(x) dx = [H(x)]_0^1 = H(1) - H(0)
= 5/ln(5) - 5/ln^2(5) - 0 + 1/ln^2(5) = 5/ln(5) - 4/ln^2(5)
= (5ln(5)-4)/ln^2(5)~~1.56.
That's it!
