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Question: Suppose that the area bounded by the curve $y = 3x^2 − 12x + 9$, the $x$-axis and the lines $x = 1$ and $x = k$ is $16$. Find $k$.

I tried but found no solution.

What I did:

$$\int_1^k (3x^2-12x+9) \,dx = \big[x^3-6x+9x \big]_1^k = k^3-6k^2+9k-4 =16$$ $$\Rightarrow k^3-6k^2+9k-20 = 0$$

Then after simplifying I got:

$$(k-5)(k^2-k+4) = 0$$

This means $k = 5$, because $(k^2-k+4)$ has no real roots, but this is wrong. The reason is after drawing the graph, the area under the curve, from $x=1$ to $x=5$ is:

$$-\int_1^3 (3x^2-12x+9) \, dx + \int_3^5 (3x^2-12x+9) \, dx = 24 \ne 16$$

Can anyone help? What mistake did I make?

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  • $\begingroup$ There is a mistake in the question. x=1 not -1 $\endgroup$ – Mohamed Nov 30 '14 at 7:23
  • $\begingroup$ Check the graph. This might help you. desmos.com/calculator/xrghc65aox . Quadratic function is passing through $x$-axis from $x=1$ and $x=3$. You'll have to break the limits accordingly. $\endgroup$ – Saharsh Nov 30 '14 at 7:24
  • $\begingroup$ the given function has two roots namely $1$ and $3$ and you have to take the absolute value of the given integral $\endgroup$ – Dr. Sonnhard Graubner Nov 30 '14 at 7:25
  • $\begingroup$ @Dr.SonnhardGraubner Thank you for answering. This is what I actually did, but instead I kept a negative sign. The actual value of the integral between 1 and 3 is -4. $\endgroup$ – Mohamed Nov 30 '14 at 7:34
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As I set the question, but do not have the reputation to comment, I will say what the question should have been, and answer it. I will also provide an answer to the question you attempted, and point out your flaw.

There was a typo on the sheet: the 1 should have been a -1 (this was corrected at some point though).

Therefore, the question goes as follows: $$ \begin{align*} 16&=\int\limits_{-1}^k3x^2-12x+9\mathrm{d}x &\text{NOTE: Do not omit the dx (as in the OP). You'll drop a mark.}\\ &=\left[x^3-6x^2+9x\right]_{-1}^k\\ &=k^3-6k^2+9k-(-1-6-9)\\ \Rightarrow 0&=k^3-6k^2+9k\\ \Rightarrow k&=0\text{ or }k=3 \end{align*} $$ We discard the $k=3$ solution so the answer is $k=0$. I will leave you to work out why we discard $k=3$ rather than $k=0$...


If we take the lower limit to be $x=1$, then note that the area between $x=1$ and $x=3$ (the root we care about) is $4$. Hence, $k>3$. Therefore, the equation you should have been solving is the following. $$ \begin{align*} 16&=-\int\limits_{-1}^33x^2-12x+9\mathrm{d}x+\int\limits_{3}^k3x^2-12x+9\mathrm{d}x\\ &=4+\int\limits_{3}^k3x^2-12x+9\mathrm{d}x \end{align*} $$ This gives you the equation $(k-3)^2k=12$. The only real root is $k\approx4.6129$.

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If you consider the area between the curve and the $x$ axis, there is not problem and you have $$A=\int_1^k (3x^2-12x+9)dx=k^3-6 k^2+9 k-4$$ as you wrote. In order to have $A=16$, then $$k^3-6 k^2+9 k-20=(k-5) (k^2-k+4)=0$$ as you correctly wrote and then the only possible solution is effectively $k=5$.

The fact that, for $1 \leq x \leq 3$, the curve is below the axis is not important. The total area is composed by a negative portion (below the axis) and a positive portion (above the axis).

They could have asked you $k$ for $A=0$ and I am sure that this would have been making less problem to you. In such a case, you would have find $k=4$.

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  • $\begingroup$ Thank you for answering, Appreciate it :) The fact that $k=4$ when $A=0$ came to my mind while solving the question. But what if Im asked to find the area bounded between the x-axis, $x=1$ and $x=5$ , shall I write $16$ or $24$ ? Or it depends on the way Im being taught? $\endgroup$ – Mohamed Nov 30 '14 at 8:34
  • $\begingroup$ You are welcome ! For this situation where one part of the curve is below the axis and another above the axis, the question has to be clarified; there is no shame to ask the question to your teacher. Just for fun : the area of a circle centered at the origin can be $0$ or $\pi r^2$ ! What does he want ? Cheers. $\endgroup$ – Claude Leibovici Nov 30 '14 at 8:42
  • $\begingroup$ Thank you for clarifying. We studied the difference between the the Integral value (which can be negative) and the area (which we need to find the absolute value for the integral below the x-axis). Anyway, thanks again for responding :) $\endgroup$ – Mohamed Nov 30 '14 at 16:24

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