Integrals anti derivatives.

Problem:  (1/3)(x + 3)²

The anti-derivative is:  (1/3)·2·(x + 3)  =  (2x + 6)/3  (by using the chain rule)

If you multiply out the original problem:  (1/3)(x + 3)²  =  (1/3)(x² + 6x + 9)

which has:  (1/3)·(2x + 6)  =  (2x + 6)/3  as its anti-derivative; the same as before.

Now:  evaluate  (2x + 6)/3  when x = 2

    and evaluate it when x = 1

    and subtract the second answer from the first.

Would you care to look at that again Geno ?

Lets see.

\(\displaystyle  \int_1^2 \frac{(x+3)^2}{3} dx\\ =\displaystyle  \frac{1}{3}\int_1^2 (x^2+6x+9) dx\\ =\displaystyle  \frac{1}{3}* \left[\frac{x^3}{3}+\frac{6x^2}{2}+9x\right] _1^2\\ =\displaystyle  \frac{1}{3}* \left[(\frac{2^3}{3}+\frac{6*2^2}{2}+9*2)-(\frac{1^3}{3}+\frac{6*1^2}{2}+9*1)\right] \\ =\displaystyle  \frac{1}{3}* \left[(\frac{8}{3}+\frac{24}{2}+18)-(\frac{1}{3}+\frac{6}{2}+9)\right] \\ =\displaystyle  \frac{1}{3}* \left[(\frac{7}{3}+\frac{18}{2}+9)\right]\\ =\displaystyle  \frac{1}{3}* \left[(\frac{7}{3}+18)\right]\\ =\displaystyle   \left[(\frac{7}{9}+6)\right]\\~\\ =6\frac{7}{9}\)

LaTex:

\displaystyle  \int_1^2 \frac{(x+3)^2}{3} dx\\
=\displaystyle  \frac{1}{3}\int_1^2 (x^2+6x+9) dx\\
=\displaystyle  \frac{1}{3}* \left[\frac{x^3}{3}+\frac{6x^2}{2}+9x\right] _1^2\\
=\displaystyle  \frac{1}{3}* \left[(\frac{2^3}{3}+\frac{6*2^2}{2}+9*2)-(\frac{1^3}{3}+\frac{6*1^2}{2}+9*1)\right] \\
=\displaystyle  \frac{1}{3}* \left[(\frac{8}{3}+\frac{24}{2}+18)-(\frac{1}{3}+\frac{6}{2}+9)\right] \\
=\displaystyle  \frac{1}{3}* \left[(\frac{7}{3}+\frac{18}{2}+9)\right]\\
=\displaystyle  \frac{1}{3}* \left[(\frac{7}{3}+18)\right]\\
=\displaystyle   \left[(\frac{7}{9}+6)\right]\\~\\
=6\frac{7}{9}