Answer:
Your answer is absolutely correct
Step-by-step explanation:
The work would be as follows:
![\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\\\\](https://tex.z-dn.net/?f=%5Cint%20_0%5E%7B%5Csqrt%7B%5Cpi%20%7D%7D4x%5E3%5Ccos%20%5Cleft%28x%5E2%5Cright%29dx%2C%5C%5C%5C%5C%5Cmathrm%7BTake%5C%3Athe%5C%3Aconstant%5C%3Aout%7D%3A%5Cquad%20%5Cint%20a%5Ccdot%20f%5Cleft%28x%5Cright%29dx%3Da%5Ccdot%20%5Cint%20f%5Cleft%28x%5Cright%29dx%5C%5C%3D%3E%204%5Ccdot%20%5Cint%20_0%5E%7B%5Csqrt%7B%5Cpi%20%7D%7Dx%5E3%5Ccos%20%5Cleft%28x%5E2%5Cright%29dx%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3Au-substitution%3A%7D%5C%3Au%3Dx%5E2%5C%5C%3D%3E%204%5Ccdot%20%5Cint%20_0%5E%7B%5Cpi%20%7D%5Cfrac%7Bu%5Ccos%20%5Cleft%28u%5Cright%29%7D%7B2%7Ddu%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3AIntegration%5C%3ABy%5C%3AParts%3A%7D%5C%3Au%3Du%2C%5C%3Av%27%3D%5Ccos%20%5Cleft%28u%5Cright%29%5C%5C%3D%3E%204%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5Bu%5Csin%20%5Cleft%28u%5Cright%29-%5Cint%20%5Csin%20%5Cleft%28u%5Cright%29du%5Cright%5D%5E%7B%5Cpi%20%7D_0%5C%5C%5C%5C)
![\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](https://tex.z-dn.net/?f=%5Cint%20%5Csin%20%5Cleft%28u%5Cright%29du%3D-%5Ccos%20%5Cleft%28u%5Cright%29%5C%5C%3D%3E%204%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5Bu%5Csin%20%5Cleft%28u%5Cright%29-%5Cleft%28-%5Ccos%20%5Cleft%28u%5Cright%29%5Cright%29%5Cright%5D%5E%7B%5Cpi%20%7D_0%5C%5C%5C%5C%5Cmathrm%7BSimplify%5C%3A%7D4%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5Bu%5Csin%20%5Cleft%28u%5Cright%29-%5Cleft%28-%5Ccos%20%5Cleft%28u%5Cright%29%5Cright%29%5Cright%5D%5E%7B%5Cpi%20%7D_0%3A%5Cquad%202%5Cleft%5Bu%5Csin%20%5Cleft%28u%5Cright%29%2B%5Ccos%20%5Cleft%28u%5Cright%29%5Cright%5D%5E%7B%5Cpi%20%7D_0%5C%5C%5C%5C%5Cmathrm%7BCompute%5C%3Athe%5C%3Aboundaries%7D%3A%5Cquad%20%5Cleft%5Bu%5Csin%20%5Cleft%28u%5Cright%29%2B%5Ccos%20%5Cleft%28u%5Cright%29%5Cright%5D%5E%7B%5Cpi%20%7D_0%3D-2%5C%5C%3D%3E%202%28-2%29%20%3D%20-%204)
Hence proved that your solution is accurate.
Here’s if you need it written out
The value is 200. It is 4 times greater
Well number 7 would be 70° because it's the same angle as the one that's labeled 70°.
Number 8 would be G, alternate exterior angles. They're not on the inside, which makes them on the outside, which is the exterior. And they are not on the same side, which makes them alternate and not corresponding.