Answer:
[tex]\int\limits^1_0 {\frac{x^{3} }{\sqrt{3+x^2} } } \, dx \approx 0.130768[/tex]
Step-by-step explanation:
Use Integration by Substitution:
For the integrand [tex]\frac{x^{3} }{\sqrt{3+x^2} }[/tex] , substitute:
[tex]u=x^2 \\du=2xdx[/tex]
Also, remember to evaluate the new integration limits:
Lower limit:
[tex]u=0^2=0[/tex]
Upper limit:
[tex]u=1^2=1[/tex]
So:
[tex]=\frac{1}{2} \int\limits^1_0 {\frac{u}{\sqrt{u+3} } } \, du[/tex]
Now, do a new substitution for the integrand [tex]\frac{u}{\sqrt{u+3} }[/tex] :
[tex]s=u+3\\ds=du[/tex]
Again, this gives us a new lower limit and a new upper limit:
Lower limit:
[tex]s=0+3=3[/tex]
Upper limit:
[tex]s=1+3=4[/tex]
[tex]=\frac{1}{2} \int\limits^4_3 {\frac{s-3}{\sqrt{s} } } \, ds[/tex]
Express [tex]\frac{s-3}{\sqrt{s} }[/tex] as [tex]\sqrt{s} - \frac{3}{\sqrt{s} }[/tex]
Now, integrate the sum term by term and factor out constants:
[tex]=\frac{1}{2} \int\limits^4_3 {\sqrt{s} } \, ds - \frac{3}{2} \int\limits^4_3 {\frac{1}{\sqrt{s} } } \, ds[/tex]
Finally, apply the fundamental theorem of calculus:
[tex]=(\frac{1}{3} s^{\frac{3}{2} } \left \{ {{4} \atop {3}} \right. ) - (3\sqrt{s} \left \{ {{4} \atop {3}} \right )=\frac{8}{3} -\sqrt{3} +3(\sqrt{3} -2)=2\sqrt{3} - \frac{10}{3} \approx0.130768[/tex]
