qual a integral 2xcos(x^2)
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É dada a integral:
Vamos fazer uma substituição. Seja
. Então: 
. Substituindo na integral dada:
![\displaystyle I=\int_{0}^{\sqrt{\frac{3\pi}{2}}} 2x\cos(x^2)\,dx\\\\
I=\int_{y_1}^{y_2} \cos(y)\,dy\\\\
I=[\sin(y)]_{y_1}^{y_2}](https://tex.z-dn.net/?f=%5Cdisplaystyle+I%3D%5Cint_%7B0%7D%5E%7B%5Csqrt%7B%5Cfrac%7B3%5Cpi%7D%7B2%7D%7D%7D+2x%5Ccos%28x%5E2%29%5C%2Cdx%5C%5C%5C%5C%0AI%3D%5Cint_%7By_1%7D%5E%7By_2%7D+%5Ccos%28y%29%5C%2Cdy%5C%5C%5C%5C%0AI%3D%5B%5Csin%28y%29%5D_%7By_1%7D%5E%7By_2%7D "\displaystyle I=\int_{0}^{\sqrt{\frac{3\pi}{2}}} 2x\cos(x^2)\,dx\\\\
I=\int_{y_1}^{y_2} \cos(y)\,dy\\\\
I=[\sin(y)]_{y_1}^{y_2}")
Voltando à variável x:
![I=[\sin(y)]_{y_1}^{y_2}\\\\
I=[\sin(x^2)]_{0}^{\sqrt{\frac{3\pi}{2}}}\\\\
I=\sin\left(\sqrt{\dfrac{3\pi}{2}}^2\right)-\sin(0^2)\\\\
I=\sin\left(\dfrac{3\pi}{2}\right)-\sin(0)\\\\
I=-1-0\\\
I=-1\\\\
\boxed{\int_{0}^{\sqrt{\frac{3\pi}{2}}} 2x\cos(x^2)\,dx=-1}](https://tex.z-dn.net/?f=I%3D%5B%5Csin%28y%29%5D_%7By_1%7D%5E%7By_2%7D%5C%5C%5C%5C%0AI%3D%5B%5Csin%28x%5E2%29%5D_%7B0%7D%5E%7B%5Csqrt%7B%5Cfrac%7B3%5Cpi%7D%7B2%7D%7D%7D%5C%5C%5C%5C%0AI%3D%5Csin%5Cleft%28%5Csqrt%7B%5Cdfrac%7B3%5Cpi%7D%7B2%7D%7D%5E2%5Cright%29-%5Csin%280%5E2%29%5C%5C%5C%5C%0AI%3D%5Csin%5Cleft%28%5Cdfrac%7B3%5Cpi%7D%7B2%7D%5Cright%29-%5Csin%280%29%5C%5C%5C%5C%0AI%3D-1-0%5C%5C%5C%0AI%3D-1%5C%5C%5C%5C%0A%5Cboxed%7B%5Cint_%7B0%7D%5E%7B%5Csqrt%7B%5Cfrac%7B3%5Cpi%7D%7B2%7D%7D%7D+2x%5Ccos%28x%5E2%29%5C%2Cdx%3D-1%7D%0A "I=[\sin(y)]_{y_1}^{y_2}\\\\
I=[\sin(x^2)]_{0}^{\sqrt{\frac{3\pi}{2}}}\\\\
I=\sin\left(\sqrt{\dfrac{3\pi}{2}}^2\right)-\sin(0^2)\\\\
I=\sin\left(\dfrac{3\pi}{2}\right)-\sin(0)\\\\
I=-1-0\\\
I=-1\\\\
\boxed{\int_{0}^{\sqrt{\frac{3\pi}{2}}} 2x\cos(x^2)\,dx=-1}")
Vamos fazer uma substituição. Seja
Voltando à variável x:
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