Answer:
[tex]=\frac{e-\sqrt[16]{e}}{4}[/tex]
Step-by-step explanation:
So we have the definite integral:
[tex]\int\limits^2_1{\frac{e^{\frac{1}{x^4}}}{x^5} \, dx[/tex]
Again, we can use u-substitution. Let u equal 1/x^4. So:
[tex]u=\frac{1}{x^4}=x^{-4}[/tex]
Find the derivative:
[tex]du=-4x^{-5}=-4(\frac{1}{x^5})dx[/tex]
Divide by -4. So du is:
[tex]-\frac{1}{4}du=\frac{1}{x^5}dx[/tex]
Of course, we also need to change our bounds. Substitute 1 and 2 into u:
[tex]u=1/(2)^4=1/16\\u=1/(1)^4=1/1=1[/tex]
Therefore, our new bounds are from 1 to 1/16.
So, make the substitutions:
[tex]=-\frac{1}{4}\int\limits^\frac{1}{16}_1 {e^u} \, du[/tex]
The integral of e^u is just e^u. So:
[tex]=-\frac{1}{4}(e^u)[/tex]
Evaluate for the bounds:
[tex]=-\frac{1}{4}(e^{\frac{1}{16}}-e^{1})[/tex]
Simplify:
[tex]=-\frac{\sqrt[16]{e}-e}{4}[/tex]
Distribute:
[tex]=\frac{e-\sqrt[16]{e}}{4}[/tex]
And we're done!
