How to calculate this integral?#int_0^1x2^xdx#

Question

Cameron Alexander · Accepted Answer

The integral equals #2/ln2 - 1/(ln2)^2 ~~ 0.804#

Use integration by parts, which states that #int u dv = uv - int v du#. We let #u = x# and #dv = 2^x#. Then #du = dx# and #v = 2^x/ln2#. For now, let's just consider the indefinite integral #int x2^xdx#.

#int x 2^x dx = x(2^x/ln2) - int 2^x/ln2 * 1 dx#

#int x2^xdx = x(2^x/ln2) - int 2^x/ln2 dx#

#int x2^xdx = (x2^x)/ln2 - 1/ln2 int 2^x dx#

#int x2^x dx = (x2^x)/ln2 - 1/ln2(2^x/ln2)#

#int x2^x dx = (x2^x)/ln2 - 2^x/(ln2)^2#

This has been calculated. We now evaluate using the second fundamental theorem of calculus which states that #int_a^b F(x) dx = f(b) - f(a)#, where #F(x)# is continuous on #[a, b]# and #f'(x) = F(x)#.

#int_0^1 x2^x dx = (1(2^1))/ln2 - 2^1/(ln2)^2 - ((0(2^0))/ln2 - 2^0/(ln2)^2)#

#int_0^1 x2^x dx = 2/ln2 - 2/(ln2)^2 + 1/(ln2)^2#

#int_0^1 x2^x dx = 2/ln2 - 1/(ln2)^2#

An estimation of this in numbers would result in

#int_0^1 x2^x dx ~~ 0.804#

I hope this is useful!

Cameron Alexander · Answer

# int \ x2^x \ dx = (2ln2 - 1)/ln^2 2 #

If you are studying mathematics, you should become familiar with the Integration By Parts (IBP) formula and practice applying it: # int \ u(dv)/dx \ dx = uv - int \ v(du)dx \ dx #, or less formally # " " int \ udv=uv-int \ vdu # "The integral of udv equals uv minus the integral of vdu" is the less formal rule that I was taught to remember. If you have trouble remembering it, try to visualize it as a direct result of integrating the Product Rule for differentiation. In essence, we want to find one function that becomes simpler when differentiated and another that becomes simpler when integrated (or at least is integrable). So for the integrand #x 2^x#, hopefully you can see that #x# simplifies when differentiated and #2^x# effectively remains unchanged under integration. Let # { (u,=x, => (du)/dx=1), ((dv)/dx,=2^x, => v=2^x/ln2 ) :}# Next, entering the IBP formula provides us with: # \ \ \ int \ (x)(2^x) \ dx = (x)(2^x/ln2) - int_0^1 \ (2^x/ln2)(1) \ dx # # :. int_0^1 \ x2^x \ dx = [(x2^x)/ln2]_0^1 - 1/ln2 \ int_0^1 \ 2^x \ dx # # " " = 1/ln2(1*2^1-0) - [1/ln2*2^x/ln2]_0^1 # # " " = 1/ln2(2) - 1/ln^2 2(2^1-2^0) # # " " = 2/ln2 - 1/ln^2 2(2-0) # # " " = 2/ln2 - 1/ln^2 2 # # " " = (2ln2 - 1)/ln^2 2 #

Charles Anctil · Answer

undefined To calculate the integral $\int_0^1 x^{2^x} dx$, there is no elementary antiderivative in terms of standard functions. Therefore, numerical methods such as approximation techniques or specialized algorithms may be employed to obtain an approximate value for this integral.