# Thread: Integrating to Find Particular Solution - Question 2

1. ## Integrating to Find Particular Solution - Question 2

The answer to the following question is C = -3.75. Please tell me where I went wrong:

Question: y = integral (squareroot 3x - cuberoot 9x+3x)dx, where x = 3 and y =9

Let U = 9x + 3x
du = 9 + 3 dx
du = 12 dx

y = integral [(3x)^2] - integral (U)^3

y = [(3x^3)/3] - [(U^4)/4] + C

y = x^3 - [(9x+3x)^4]/4 + C

9 = 3^3 - [(9{3}+3{3})^4]/4 + C

9 = 27 - [(27+9)^4]/4 + C

9 = 27 - 1679616/4 + C

C = 419886 (According to my teacher, the answer is C = -3.75)

2. Originally Posted by sparky
The answer to the following question is C = -3.75. Please tell me where I went wrong:

Question: y = integral (squareroot 3x - cuberoot 9x+3x)dx, where x = 3 and y =9

Let U = 9x + 3x
du = 9 + 3 dx
du = 12 dx

y = integral [(3x)^2] - integral (U)^3

y = [(3x^3)/3] - [(U^4)/4] + C

y = x^3 - [(9x+3x)^4]/4 + C

9 = 3^3 - [(9{3}+3{3})^4]/4 + C

9 = 27 - [(27+9)^4]/4 + C

9 = 27 - 1679616/4 + C

C = 419886 (According to my teacher, the answer is C = -3.75)
$y = \int \left ( \sqrt{3x} - \sqrt[3]{3x + 9x} \right )~dx$

Then your first term changes to
$y = \int (3x)^2 - ~...$

(Edit: You did the same thing with the cube root.)

Also, I am suspicious about that cube root term. Is there a typo? I can't think of the reason why it wasn't given as 12x in the first place?

-Dan

3. That's the question: y = integral (square root of 3x minus cube root of (3x + 9x))

Maybe the 3x + 9x was put there to confuse me. Where did I go wrong in my working?

4. Originally Posted by sparky
That's the question: y = integral (square root of 3x minus cube root of (3x + 9x))

Maybe the 3x + 9x was put there to confuse me. Where did I go wrong in my working?
I have not completed the problem myself, but as I said, you changed $\sqrt{3x}$ into $(3x)^2$ and $\sqrt[3]{3x + 9x}$ into $(3x + 9x)^3$.

Try fixing that and see what happens.

-Dan

5. Ok, I tried it again and got it wrong again:

y = integral (square root of 3x minus cuberoot of (9x + 3x))dx

y = integral of (3x)^2 - integral of (9x)^3 + integral of (3x)^3

y = [(3x)^3]/3 - [(9x)^4]/4 + [(3x)^4]/4 + C

y = x^3 - [(12x)^4]/4 + C

y = x^3 - 3x^4 + C

9 = 3^3 - (3{3}^4) + C

9 = 27 - 243 + C

C = 225

What am I doing wrong?

6. Originally Posted by sparky
Ok, I tried it again and got it wrong again:

y = integral (square root of 3x minus cuberoot of (9x + 3x))dx

y = integral of (3x)^2 - integral of (9x)^3 + integral of (3x)^3
What am I doing wrong?
$\int \left ( \sqrt{3x} - \sqrt[3]{3x + 9x} \right )~dx$

or should it be
$\int \left ( (3x)^2 - (3x + 9x)^3 \right )~dx$

They are NOT the same!!

-Dan

Edit: And
$(9x + 3x)^3\neq (9x)^3 + (3x)^3$

$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

Edit II:
I am suspecting that the integral is written incorrectly. Neither form of the problem I listed above gives you the given answer of c = -3.75. The most likely source of error is the 9x + 3x term. I suspect it has been miscopied either by you or the source you got the problem from.