математика помогите!!!
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Ответил 21WAR
0
a)
![y=-x^{2}+4x y=0 x=2 \
-x^{2}+4x=0 \
-x(x-4)=0 \
x=0 x= 4 \
S=int limits^0_2(-x^{2}+4x)dx = mid^0_2(-frac{x^{3}}{3}+2x^{2}) \
-frac{8}{3}+8=-frac{16}{3}](https://tex.z-dn.net/?f=y%3D-x%5E%7B2%7D%2B4x++y%3D0++x%3D2+%5C%0A-x%5E%7B2%7D%2B4x%3D0+%5C%0A-x%28x-4%29%3D0+%5C%0Ax%3D0++x%3D+4+%5C%0AS%3Dint+limits%5E0_2%28-x%5E%7B2%7D%2B4x%29dx+%3D+mid%5E0_2%28-frac%7Bx%5E%7B3%7D%7D%7B3%7D%2B2x%5E%7B2%7D%29+%5C%0A-frac%7B8%7D%7B3%7D%2B8%3D-frac%7B16%7D%7B3%7D+ "y=-x^{2}+4x y=0 x=2 \
-x^{2}+4x=0 \
-x(x-4)=0 \
x=0 x= 4 \
S=int limits^0_2(-x^{2}+4x)dx = mid^0_2(-frac{x^{3}}{3}+2x^{2}) \
-frac{8}{3}+8=-frac{16}{3}")
b)
![y=cosx \
y=0 x=-frac{pi}{6} x=frac{pi}{6} \
S=int limits^frac{-pi}{6}_frac{pi}{6}(cosx)dx \
mid^frac{-pi}{6}_frac{pi}{6}(-sinx)=mid^frac{5pi}{6}_frac{pi}{6}-sinx \
-frac{1}{2}-frac{1}{2}=-1](https://tex.z-dn.net/?f=y%3Dcosx+%5C%0Ay%3D0++x%3D-frac%7Bpi%7D%7B6%7D++x%3Dfrac%7Bpi%7D%7B6%7D+%5C%0AS%3Dint+limits%5Efrac%7B-pi%7D%7B6%7D_frac%7Bpi%7D%7B6%7D%28cosx%29dx+%5C+%0Amid%5Efrac%7B-pi%7D%7B6%7D_frac%7Bpi%7D%7B6%7D%28-sinx%29%3Dmid%5Efrac%7B5pi%7D%7B6%7D_frac%7Bpi%7D%7B6%7D-sinx+%5C%0A-frac%7B1%7D%7B2%7D-frac%7B1%7D%7B2%7D%3D-1+ "y=cosx \
y=0 x=-frac{pi}{6} x=frac{pi}{6} \
S=int limits^frac{-pi}{6}_frac{pi}{6}(cosx)dx \
mid^frac{-pi}{6}_frac{pi}{6}(-sinx)=mid^frac{5pi}{6}_frac{pi}{6}-sinx \
-frac{1}{2}-frac{1}{2}=-1")
v)
![y=x^{2} y=x+2 \
x^{2}=x+2 \
x^{2}-x-2=0 \
x_{1}=2 x_{2}=-1 \
S=intlimits^2_1=mid^1_2 frac{x^{2}}{2}+2x=frac{-3}{2}+6=-frac{15}{2} \
S=-3+frac{15}{2}=frac{9}{2}](https://tex.z-dn.net/?f=y%3Dx%5E%7B2%7D++y%3Dx%2B2+%5C%0Ax%5E%7B2%7D%3Dx%2B2+%5C%0Ax%5E%7B2%7D-x-2%3D0+%5C%0Ax_%7B1%7D%3D2++x_%7B2%7D%3D-1+%5C%0AS%3Dintlimits%5E2_1%3Dmid%5E1_2+frac%7Bx%5E%7B2%7D%7D%7B2%7D%2B2x%3Dfrac%7B-3%7D%7B2%7D%2B6%3D-frac%7B15%7D%7B2%7D+%5C%0AS%3D-3%2Bfrac%7B15%7D%7B2%7D%3Dfrac%7B9%7D%7B2%7D "y=x^{2} y=x+2 \
x^{2}=x+2 \
x^{2}-x-2=0 \
x_{1}=2 x_{2}=-1 \
S=intlimits^2_1=mid^1_2 frac{x^{2}}{2}+2x=frac{-3}{2}+6=-frac{15}{2} \
S=-3+frac{15}{2}=frac{9}{2}")
b)
v)
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