Considera la función 
Solución:
Para esbozar la gráfica de 
Luego, y=-mx es una recta decreciente que corta a la parábola en:
El área del recinto sombreado S ha de ser 36.
![\displaystyle S=\int_0^{2m}(-x^2+mx)-(-mx)~dx=\int_0^{2m}-x^2+2mx~dx=\\\\=\left[\frac{-x^3}3+\frac{2mx^2}2\right]_0^{2m}=\left[\frac{-x^3+3mx^2}3\right]_0^{2m}=\\\\=\left(\frac{-(2m)^3+3m(2m)^2}3\right)-\left(\frac{-0^3+2m\cdot 0^2}3\right)=\\\\=\frac{-8m^3+12m^3}3=\frac{4m^3}3](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+S%3D%5Cint_0%5E%7B2m%7D%28-x%5E2%2Bmx%29-%28-mx%29%7Edx%3D%5Cint_0%5E%7B2m%7D-x%5E2%2B2mx%7Edx%3D%5C%5C%5C%5C%3D%5Cleft%5B%5Cfrac%7B-x%5E3%7D3%2B%5Cfrac%7B2mx%5E2%7D2%5Cright%5D_0%5E%7B2m%7D%3D%5Cleft%5B%5Cfrac%7B-x%5E3%2B3mx%5E2%7D3%5Cright%5D_0%5E%7B2m%7D%3D%5C%5C%5C%5C%3D%5Cleft%28%5Cfrac%7B-%282m%29%5E3%2B3m%282m%29%5E2%7D3%5Cright%29-%5Cleft%28%5Cfrac%7B-0%5E3%2B2m%5Ccdot+0%5E2%7D3%5Cright%29%3D%5C%5C%5C%5C%3D%5Cfrac%7B-8m%5E3%2B12m%5E3%7D3%3D%5Cfrac%7B4m%5E3%7D3&bg=ffffff&fg=141412&s=0&c=20201002 "\displaystyle S=\int_0^{2m}(-x^2+mx)-(-mx)~dx=\int_0^{2m}-x^2+2mx~dx=\\\\=\left[\frac{-x^3}3+\frac{2mx^2}2\right]_0^{2m}=\left[\frac{-x^3+3mx^2}3\right]_0^{2m}=\\\\=\left(\frac{-(2m)^3+3m(2m)^2}3\right)-\left(\frac{-0^3+2m\cdot 0^2}3\right)=\\\\=\frac{-8m^3+12m^3}3=\frac{4m^3}3")
Igualamos a 36 y resolvemos:
![\displaystyle \frac{4m^3}3=36\\\\4m^3=108\\\\m^3=27\\\\m=\sqrt[3]27=3](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B4m%5E3%7D3%3D36%5C%5C%5C%5C4m%5E3%3D108%5C%5C%5C%5Cm%5E3%3D27%5C%5C%5C%5Cm%3D%5Csqrt%5B3%5D27%3D3&bg=ffffff&fg=141412&s=0&c=20201002 "\displaystyle \frac{4m^3}3=36\\\\4m^3=108\\\\m^3=27\\\\m=\sqrt[3]27=3")
♦
Considera la función
Solución:
Para esbozar la gráfica de
Luego, y=-mx es una recta decreciente que corta a la parábola en:
El área del recinto sombreado S ha de ser 36.
Igualamos a 36 y resolvemos:
♦
