Integral Example - # 4

[Jason76]

\(\displaystyle \int 5x(x^{2} + 4)^{2} dx\)

\(\displaystyle 5x (u)^{2} dx\)

\(\displaystyle u = x^{2} + 4\)

\(\displaystyle du = 2x\)

\(\displaystyle \dfrac{5}{2} du = \dfrac{5}{2} 2x\)

\(\displaystyle du = 5x\)

\(\displaystyle \dfrac{5}{2}\int u^{2} dx\)

\(\displaystyle \rightarrow \dfrac{5}{2} \dfrac{u^{3}}{3} + C\)

\(\displaystyle \rightarrow \dfrac{5u^{3}}{6} + C\)

\(\displaystyle \rightarrow \dfrac{5(x^{2} + 4)^{3}}{6} + C\) The computer says this is wrong.

The computer's answer is \(\displaystyle \dfrac{5x^{6}}{6} + 10x^{4} + 40x^{2} + C\)

 

[stapel]

\(\displaystyle \int 5x(x^{2} + 4)^{2} dx\)

\(\displaystyle 5x (u)^{2} dx\)

\(\displaystyle u = x^{2} + 4\)

\(\displaystyle du = 2x\)

No; du = 2x dx.

\(\displaystyle \dfrac{5}{2} du = \dfrac{5}{2} 2x\)

\(\displaystyle du = 5x\)

No; (5/2)du = (5x)dx.

\(\displaystyle \rightarrow \dfrac{5(x^{2} + 4)^{3}}{6} + C\) The computer says this is wrong.

The computer's answer is \(\displaystyle \dfrac{5x^{6}}{6} + 10x^{4} + 40x^{2} + C\)

We can't help how your computer is expecting the answer to be formatted. What efforts have you made to determine how your expression expands? What comparisons have you made between your expression and the desired form?

What, precisely, is your question?

 

[JeffM]

\(\displaystyle \int 5x(x^{2} + 4)^{2} dx\)

\(\displaystyle 5x (u)^{2} dx\)

\(\displaystyle u = x^{2} + 4\)

\(\displaystyle du = 2x\) No! \(\displaystyle du = 2xdx.\)

\(\displaystyle \dfrac{5}{2} du = \dfrac{5}{2} 2x\)\(\displaystyle dx\)

\(\displaystyle du = 5x\) Not at all. \(\displaystyle \dfrac{5}{2}du = 5xdx\)

\(\displaystyle \dfrac{5}{2}\int u^{2} dx\) NO! NO!!! \(\displaystyle \dfrac{5}{2}\int u^{2} du\)

\(\displaystyle \rightarrow \dfrac{5}{2} \dfrac{u^{3}}{3} + C\)

\(\displaystyle \rightarrow \dfrac{5u^{3}}{6} + C\)

\(\displaystyle \rightarrow \dfrac{5(x^{2} + 4)^{3}}{6} + C\) The computer says this is wrong.

The computer's answer is \(\displaystyle \dfrac{5x^{6}}{6} + 10x^{4} + 40x^{2} + C\)

\(\displaystyle \dfrac{5(x^2 + 4)^3}{6} + K = \)

\(\displaystyle \dfrac{5(x^2 + 4)^2(x^2 + 4)}{6} + K =\)

\(\displaystyle \dfrac{5(x^4 + 8x^2 + 16)(x^2 + 4)}{6} + K =\)

\(\displaystyle \dfrac{5x^2(x^4 + 8x^2 + 16) + 5 * 4(x^4 + 8x^2 + 16)}{6} + K =\)

\(\displaystyle \dfrac{5x^6 + 40x^4 + 80x^2 + 20x^4 + 160x^2 + 320}{6} + K =\)

\(\displaystyle \dfrac{5x^6 + 60x^4 + 240x^2 + 320}{6} + K =\)

\(\displaystyle \dfrac{5x^6}{6} + 10x^4 + 40x^2 + C,\ where\ C = \dfrac{320}{6} + K.\)

Just algebra.

 

[Jason76]

Oh i see, it was done right, but just needed factoring out.

\(\displaystyle \int 5x(x^{2} + 4)^{2} dx\)

\(\displaystyle 5x (u)^{2} dx\)

\(\displaystyle u = x^{2} + 4\)

\(\displaystyle du = 2x dx\)

\(\displaystyle \dfrac{5}{2} du = \dfrac{5}{2} 2x dx\) notation corrected

\(\displaystyle du = 5x dx\)

\(\displaystyle \dfrac{5}{2}\int u^{2} du\)

\(\displaystyle \rightarrow \dfrac{5}{2} \dfrac{u^{3}}{3} + C\)

\(\displaystyle \rightarrow \dfrac{5u^{3}}{6} + C\)

\(\displaystyle \rightarrow \dfrac{5(x^{2} + 4)^{3}}{6} + C\)

And then factoring out....

 

[stapel]

Oh i see, it was done right, but just needed factoring out.

No; your form already is factored. As mentioned previously, you need now to multiply out, simplify, and compare.

 

[Jason76]

Factoring out at the start and completing the problem will work, but also you can do substitution, and then factor the answer. Both ways lead to the same answer. Since the factoring is a little work, neither way is easier.

 

[srmichael]

Factoring out at the start...

UGH!!!! NO!!!! Similar to what stapel said above, your function is ALREADY factored out. You mean you could have MULTIPLIED it out and simplifed then taken the integral of each resulting term.

Why does it seem, Jason, that you do not even bother to read the comments we post over these myriad of problems you have posted as you consistently make the same mistakes over and over and over and....