Найдите площадь фигуры, ограниченной параболами у=6x^{2} , у=(х-3)(х-4) и осью Ох
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Для начала находим абсциссы точек пересечения заданных графиков:
![6x^2=(x-3)(x-4)
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6x^2=x^2-7x+12
\
5x^2+7x-12=0
\
D=7^2+4cdot5cdot12=289
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x_1= frac{-7-17}{10} =-2.4
\
x_2= frac{-7+17}{10} =1](https://tex.z-dn.net/?f=6x%5E2%3D%28x-3%29%28x-4%29%0A%5C%0A6x%5E2%3Dx%5E2-7x%2B12%0A%5C%0A5x%5E2%2B7x-12%3D0%0A%5C%0AD%3D7%5E2%2B4cdot5cdot12%3D289%0A%5C%0Ax_1%3D+frac%7B-7-17%7D%7B10%7D+%3D-2.4%0A%5C%0Ax_2%3D+frac%7B-7%2B17%7D%7B10%7D+%3D1 "6x^2=(x-3)(x-4)
\
6x^2=x^2-7x+12
\
5x^2+7x-12=0
\
D=7^2+4cdot5cdot12=289
\
x_1= frac{-7-17}{10} =-2.4
\
x_2= frac{-7+17}{10} =1")
![intlimits^1_{-2.4}(x^2-7x+12-6x^2)=intlimits^1_{-2.4}(12-7x-5x^2)=
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=(12x-frac{7x^2}{2}-frac{5x^3}{3})^1_{-2.4}=
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=(12cdot1-frac{7cdot1^2}{2}-frac{5cdot1^3}{3})-(12cdot(-2.4)-frac{7cdot(-2.4)^2}{2}-frac{5cdot(-2.4)^3}{3})=
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=(12- frac{7}{2}-frac{5}{3})-(-28.8-frac{504}{25}+frac{576}{25})=
-frac{21}{6}-frac{10}{6}+40.8-frac{72}{25}=
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=-frac{31}{6}+40.8-2.88=-5frac{1}{6}+37.92=32 frac{92}{100}-frac{1}{6}=32frac{138}{150}-frac{25}{150}=32frac{113}{150}](https://tex.z-dn.net/?f=intlimits%5E1_%7B-2.4%7D%28x%5E2-7x%2B12-6x%5E2%29%3Dintlimits%5E1_%7B-2.4%7D%2812-7x-5x%5E2%29%3D%0A%5C%0A%3D%2812x-frac%7B7x%5E2%7D%7B2%7D-frac%7B5x%5E3%7D%7B3%7D%29%5E1_%7B-2.4%7D%3D%0A%5C%0A%3D%2812cdot1-frac%7B7cdot1%5E2%7D%7B2%7D-frac%7B5cdot1%5E3%7D%7B3%7D%29-%2812cdot%28-2.4%29-frac%7B7cdot%28-2.4%29%5E2%7D%7B2%7D-frac%7B5cdot%28-2.4%29%5E3%7D%7B3%7D%29%3D%0A%5C%0A%3D%2812-+frac%7B7%7D%7B2%7D-frac%7B5%7D%7B3%7D%29-%28-28.8-frac%7B504%7D%7B25%7D%2Bfrac%7B576%7D%7B25%7D%29%3D%0A-frac%7B21%7D%7B6%7D-frac%7B10%7D%7B6%7D%2B40.8-frac%7B72%7D%7B25%7D%3D%0A%5C%0A%3D-frac%7B31%7D%7B6%7D%2B40.8-2.88%3D-5frac%7B1%7D%7B6%7D%2B37.92%3D32+frac%7B92%7D%7B100%7D-frac%7B1%7D%7B6%7D%3D32frac%7B138%7D%7B150%7D-frac%7B25%7D%7B150%7D%3D32frac%7B113%7D%7B150%7D "intlimits^1_{-2.4}(x^2-7x+12-6x^2)=intlimits^1_{-2.4}(12-7x-5x^2)=
\
=(12x-frac{7x^2}{2}-frac{5x^3}{3})^1_{-2.4}=
\
=(12cdot1-frac{7cdot1^2}{2}-frac{5cdot1^3}{3})-(12cdot(-2.4)-frac{7cdot(-2.4)^2}{2}-frac{5cdot(-2.4)^3}{3})=
\
=(12- frac{7}{2}-frac{5}{3})-(-28.8-frac{504}{25}+frac{576}{25})=
-frac{21}{6}-frac{10}{6}+40.8-frac{72}{25}=
\
=-frac{31}{6}+40.8-2.88=-5frac{1}{6}+37.92=32 frac{92}{100}-frac{1}{6}=32frac{138}{150}-frac{25}{150}=32frac{113}{150}")
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