Se sabe que la función 
es continua.
a) Determina a.
b) Para a=8, calcula 
Solución:
a) Estudiemos la continuidad en x=8:
Para que sea continua ha de ser
b) Para a=8:
![\displaystyle \int_0^{10}f(x)~dx=\int_0^8\sqrt{8x}~dx+\int_8^{10}\dfrac{x^2-32}{x-4}~dx=\\\\=\int_0^8\sqrt{8x}~dx+\int_8^{10}x+4~dx+\int_8^{10}\frac{-16}{x-4}~dx=\\\\=\sqrt8\int_0^8x^{\frac 12}~dx+\left[\frac{x^2}2+4x\right]_8^{10}-16[\ln|x-4|]_8^{10}=\\\\=\sqrt{8}\left[\dfrac{x^{\frac 32}}{\frac 32}\right]_0^8+(50+40)-(32+32)-16(\ln(6)-\ln(4))=\\\\=\dfrac{2\sqrt{8}}3(\sqrt{8^3})+90-64-16\ln\left(\frac 64\right)=\\\\=\dfrac{128}3+26-16\ln\left(\frac 32\right)=\frac{206}3-16\ln\left(\frac 32\right)\approx 62.18](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cint_0%5E%7B10%7Df%28x%29%7Edx%3D%5Cint_0%5E8%5Csqrt%7B8x%7D%7Edx%2B%5Cint_8%5E%7B10%7D%5Cdfrac%7Bx%5E2-32%7D%7Bx-4%7D%7Edx%3D%5C%5C%5C%5C%3D%5Cint_0%5E8%5Csqrt%7B8x%7D%7Edx%2B%5Cint_8%5E%7B10%7Dx%2B4%7Edx%2B%5Cint_8%5E%7B10%7D%5Cfrac%7B-16%7D%7Bx-4%7D%7Edx%3D%5C%5C%5C%5C%3D%5Csqrt8%5Cint_0%5E8x%5E%7B%5Cfrac+12%7D%7Edx%2B%5Cleft%5B%5Cfrac%7Bx%5E2%7D2%2B4x%5Cright%5D_8%5E%7B10%7D-16%5B%5Cln%7Cx-4%7C%5D_8%5E%7B10%7D%3D%5C%5C%5C%5C%3D%5Csqrt%7B8%7D%5Cleft%5B%5Cdfrac%7Bx%5E%7B%5Cfrac+32%7D%7D%7B%5Cfrac+32%7D%5Cright%5D_0%5E8%2B%2850%2B40%29-%2832%2B32%29-16%28%5Cln%286%29-%5Cln%284%29%29%3D%5C%5C%5C%5C%3D%5Cdfrac%7B2%5Csqrt%7B8%7D%7D3%28%5Csqrt%7B8%5E3%7D%29%2B90-64-16%5Cln%5Cleft%28%5Cfrac+64%5Cright%29%3D%5C%5C%5C%5C%3D%5Cdfrac%7B128%7D3%2B26-16%5Cln%5Cleft%28%5Cfrac+32%5Cright%29%3D%5Cfrac%7B206%7D3-16%5Cln%5Cleft%28%5Cfrac+32%5Cright%29%5Capprox+62.18&bg=ffffff&fg=141412&s=0 "\displaystyle \int_0^{10}f(x)~dx=\int_0^8\sqrt{8x}~dx+\int_8^{10}\dfrac{x^2-32}{x-4}~dx=\\\\=\int_0^8\sqrt{8x}~dx+\int_8^{10}x+4~dx+\int_8^{10}\frac{-16}{x-4}~dx=\\\\=\sqrt8\int_0^8x^{\frac 12}~dx+\left[\frac{x^2}2+4x\right]_8^{10}-16[\ln|x-4|]_8^{10}=\\\\=\sqrt{8}\left[\dfrac{x^{\frac 32}}{\frac 32}\right]_0^8+(50+40)-(32+32)-16(\ln(6)-\ln(4))=\\\\=\dfrac{2\sqrt{8}}3(\sqrt{8^3})+90-64-16\ln\left(\frac 64\right)=\\\\=\dfrac{128}3+26-16\ln\left(\frac 32\right)=\frac{206}3-16\ln\left(\frac 32\right)\approx 62.18")
♦
Se sabe que la función
es continua.
a) Determina a.
b) Para a=8, calcula
Solución:
a) Estudiemos la continuidad en x=8:
Para que sea continua ha de ser
b) Para a=8:
♦
