Sea la función 
a) Su gráfica determina con el eje de abscisas un recinto limitado D. Calcula su área.
b) La gráfica de la función 
c) Calcular el área del recinto 
Solución:
a) f es una función cuadrática cuya gráfica es una parábola concava como 
El área D sombreada vale:
![\displaystyle D=\int_{-2}^24-x^2~dx=\left[4x-\dfrac{x^3}3\right]_{-2}^2=\\\\=\left(4\cdot2-\dfrac{2^3}3\right)-\left(4\cdot(-2)-\dfrac{(-2)^3}3\right)=\\\\=8-\dfrac83+8-\dfrac83=\boxed{\dfrac{32}3\text{ u.a.}}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+D%3D%5Cint_%7B-2%7D%5E24-x%5E2%7Edx%3D%5Cleft%5B4x-%5Cdfrac%7Bx%5E3%7D3%5Cright%5D_%7B-2%7D%5E2%3D%5C%5C%5C%5C%3D%5Cleft%284%5Ccdot2-%5Cdfrac%7B2%5E3%7D3%5Cright%29-%5Cleft%284%5Ccdot%28-2%29-%5Cdfrac%7B%28-2%29%5E3%7D3%5Cright%29%3D%5C%5C%5C%5C%3D8-%5Cdfrac83%2B8-%5Cdfrac83%3D%5Cboxed%7B%5Cdfrac%7B32%7D3%5Ctext%7B+u.a.%7D%7D&bg=ffffff&fg=141412&s=0&c=20201002 "\displaystyle D=\int_{-2}^24-x^2~dx=\left[4x-\dfrac{x^3}3\right]_{-2}^2=\\\\=\left(4\cdot2-\dfrac{2^3}3\right)-\left(4\cdot(-2)-\dfrac{(-2)^3}3\right)=\\\\=8-\dfrac83+8-\dfrac83=\boxed{\dfrac{32}3\text{ u.a.}}")
b) La función 
c) Nos piden calcular el área de la región
Luego, el área
![\displaystyle D_2=\int_{-1}^1(4-x^2)-3x^2~dx=\int_{-1}^14-4x^2~dx=\\\\=\left[4x-\dfrac{4x^3}3\right]_{-1}^1=\\\\=\left(4\cdot1-\dfrac{4\cdot1^3}3\right)-\left(4\cdot(-1)-\dfrac{4\cdot(-1)^3}3\right)=\\\\=4-\dfrac43+4-\dfrac43=8-\dfrac83=\boxed{\dfrac{16}3\text{ u.a.}}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+D_2%3D%5Cint_%7B-1%7D%5E1%284-x%5E2%29-3x%5E2%7Edx%3D%5Cint_%7B-1%7D%5E14-4x%5E2%7Edx%3D%5C%5C%5C%5C%3D%5Cleft%5B4x-%5Cdfrac%7B4x%5E3%7D3%5Cright%5D_%7B-1%7D%5E1%3D%5C%5C%5C%5C%3D%5Cleft%284%5Ccdot1-%5Cdfrac%7B4%5Ccdot1%5E3%7D3%5Cright%29-%5Cleft%284%5Ccdot%28-1%29-%5Cdfrac%7B4%5Ccdot%28-1%29%5E3%7D3%5Cright%29%3D%5C%5C%5C%5C%3D4-%5Cdfrac43%2B4-%5Cdfrac43%3D8-%5Cdfrac83%3D%5Cboxed%7B%5Cdfrac%7B16%7D3%5Ctext%7B+u.a.%7D%7D&bg=ffffff&fg=141412&s=0&c=20201002 "\displaystyle D_2=\int_{-1}^1(4-x^2)-3x^2~dx=\int_{-1}^14-4x^2~dx=\\\\=\left[4x-\dfrac{4x^3}3\right]_{-1}^1=\\\\=\left(4\cdot1-\dfrac{4\cdot1^3}3\right)-\left(4\cdot(-1)-\dfrac{4\cdot(-1)^3}3\right)=\\\\=4-\dfrac43+4-\dfrac43=8-\dfrac83=\boxed{\dfrac{16}3\text{ u.a.}}")
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