Question

∫0^{3}(3+2t^{2}𝛿(t-2)dt

Answer 1

$\int _0^3\left(3+2t^2\right)\left(t-2\right)dt\\ =\int _0^3\frac{9-4t^4}{3-2t^2}\left(t-2\right)dt\\ \mathrm{If\:exist}\:b,\:a<b<c,\:f\left(b\right)=\mathrm{undefined},\:\int _a^c\:f\left(x\right)dx=\int _a^b\:f\left(x\right)dx+\int _b^c\:f\left(x\right)dx\\ =\int _0^{\sqrt{\frac{3}{2}}}\frac{9-4t^4}{3-2t^2}\left(t-2\right)dt+\int _{\sqrt{\frac{3}{2}}}^3\frac{9-4t^4}{3-2t^2}\left(t-2\right)dt\\ =-4\sqrt{6}+\frac{27}{8}\\ =4\sqrt{6}-\frac{27}{8}\\ =-4\sqrt{6}+\frac{27}{8}+4\sqrt{6}-\frac{27}{8}\\ =0$