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∫((x+3)^2)/3 dx = ?

 1

 

 

please I can't figure it out

 Apr 17, 2022
 #3
avatar+23183 
+1

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.

 Apr 17, 2022
 #4
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Would you care to look at that again Geno ?

 Apr 18, 2022
 #5
avatar+117236 
+1

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}

 Apr 18, 2022
edited by Melody  Apr 18, 2022

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