Answer:
Your answer is absolutely correct
Step-by-step explanation:
The work would be as follows:
[tex]\int _0^{\sqrt{\pi }}4x^3\cos \left(x^2\right)dx,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=> 4\cdot \int _0^{\sqrt{\pi }}x^3\cos \left(x^2\right)dx\\\\\mathrm{Apply\:u-substitution:}\:u=x^2\\=> 4\cdot \int _0^{\pi }\frac{u\cos \left(u\right)}{2}du\\\\\mathrm{Apply\:Integration\:By\:Parts:}\:u=u,\:v'=\cos \left(u\right)\\=> 4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\int \sin \left(u\right)du\right]^{\pi }_0\\\\[/tex]
[tex]\int \sin \left(u\right)du=-\cos \left(u\right)\\=> 4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\left(-\cos \left(u\right)\right)\right]^{\pi }_0\\\\\mathrm{Simplify\:}4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\left(-\cos \left(u\right)\right)\right]^{\pi }_0:\quad 2\left[u\sin \left(u\right)+\cos \left(u\right)\right]^{\pi }_0\\\\\mathrm{Compute\:the\:boundaries}:\quad \left[u\sin \left(u\right)+\cos \left(u\right)\right]^{\pi }_0=-2\\=> 2(-2) = - 4[/tex]
Hence proved that your solution is accurate.
Answer:
[tex]\int\limits^{\sqrt{\pi}}_0 {4x^3\cos(x^2)} \, dx=-4[/tex]
Step-by-step explanation:
So we have the integral:
[tex]\int\limits^{\sqrt{\pi}}_0 {4x^3\cos(x^2)} \, dx[/tex]
As told, let's use u-substitution first and then use integration by parts.
For the u-substitution, we can let u to be equal to x². So:
[tex]u=x^2[/tex]
Differentiate:
[tex]du=2x\, dx[/tex]
We can rewrite our integral as:
[tex]\int\limits^{\sqrt{\pi}}_0 {2x(2x^2)\cos(x^2)} \, dx[/tex]
Therefore, by making our u-substitution, our integral is now:
[tex]\int\limits {2u\cos(u)} \, du[/tex]
We also need to change our bounds. Substitute them into u. So:
[tex]u=\sqrt{\pi}^2=\pi\\u=(0)^2=0[/tex]
Therefore, our integral with our new bounds is:
[tex]\int\limits^{\pi}_{0} {2u\cos(u)} \, du[/tex]
Now, let's use integration by parts. Integration by parts is given by:
[tex]\int\limits {v}\, dy=vy-\int y\, dv[/tex]
(I changed the standard u to y because we are already using u).
Let's let v be 2u and let's let dy be cos(u). Thus:
[tex]v=2u\\dv=2\,du[/tex]
And:
[tex]dy=\cos(u)\\y=\sin(u)[/tex]
So, do integration by parts:
[tex]=2u\sin(u)-\int \sin(u)2\,du[/tex]
Simplify:
[tex]=2u\sin(u)-2\int \sin(u)\,du[/tex]
Evaluate the integral:
[tex]=2u\sin(u)+2\cos(u)[/tex]
Now, use the bounds. So:
[tex](2(\pi)\sin(\pi)+2\cos(\pi))-(2(0)\sin(0)+2\cos(0))[/tex]
Evaluate:
[tex]=(2\pi(0)+2(-1))-(0(0)+2(1))[/tex]
Simplify:
[tex]=(-2)-(2)[/tex]
Subtract:
[tex]=-4[/tex]
And we're done!
