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I'm currently doing a simple integration question:

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Here is my working/solution so far:

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I have calculated this several times and only be seem to be getting a negative number as the final result. I know this is wrong as it is an area that needs to be calculated and therefore cannot be a negative number.

Any help on this is very much appreciated. Thank you!

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    $\begingroup$ I haven't checked your work yet, but why can't an integral be negative? All that means is that the next area is below the $x$-axis, but that isn't forbidden. $\endgroup$ – Ahaan S. Rungta Aug 14 '13 at 19:30
  • $\begingroup$ Plot the curve you are integrating; that should be a guide to your result. $\endgroup$ – Ron Gordon Aug 14 '13 at 19:32
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It seems your work is in an image which is appearing weirdly on my screen. I'll just outline my work.

$ \displaystyle\int \left( 5 + \dfrac{5}{4\sqrt{x}} - x^4 \right) \, \mathrm{d}x = \displaystyle\int 5 \, \mathrm{d}x + \displaystyle\int 4 \cdot x^{\frac{1}{2}} \, \mathrm{d}x - \displaystyle\int x^4 \, \mathrm{d}x $

$ = 5x + \dfrac {8}{3} x^{\frac{3}{2}} - \dfrac {x^5}{5} + \mathcal{C} $

Now, apply the limits of integration. At $ x = 3 $, the function value is $ \approx -19.7 $. At $ x = 2 $, the function value is $ \approx 11.1 $. The answer is, thus: $$ \approx -19.7 - 11.1 = \boxed {-30.8}. $$

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  • $\begingroup$ Wouldn't it be $$4x^{-1/2}dx$$ instead of $$4x^{1/2}dx$$? $\endgroup$ – hax0r_n_code Aug 15 '13 at 2:11
  • $\begingroup$ Yes, thanks a lot! I can't think. I'm not going to edit my work, but yes, you're right. Sorry about that! $\endgroup$ – Ahaan S. Rungta Aug 15 '13 at 15:39
  • $\begingroup$ No worries! Happy to help :-) $\endgroup$ – hax0r_n_code Aug 15 '13 at 22:33
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The entire curve, as you can readily check, is below the $x$-axis from $2$ to $3$. In general, the area of the region between the curves $y=f(x)$ and $y=g(x)$, from $x=a$ to $x=b$ is $$\int_a^b|f(x)-g(x)|\,dx.$$

Another version of the same thing is that the area below $y=f(x)$ and above $g(x)$ is the integral over the appropriate interval of $f(x)-g(x)$. In your case, the upper curve is $y=0$, the lower one is $y=g(x)$, where $g(x)$ is your complicated expression. So you want to integrate $0-g(x)$, or more simply $-g(x)$.

In your case, since the curve does not shift from above to below, you can take the absolute value at the end. But technically you should have been integrating the absolute value of your complicated function, which is the sum with signs reversed.

Remark: Some people would consider the negative answer perfectly reasonable, because the region is below the $x$-axis. It is a matter of convention. I don't like it. We seem to agree that area should be non-negative.

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  • $\begingroup$ It's not really a convention, but a consequence of the definition of Riemann Sums. $\endgroup$ – FireGarden Aug 15 '13 at 12:06
  • $\begingroup$ The definition of integral is not at issue, the integral of the given function from $2$ to $3$ is negative. However, the question asks for area. Area is in particular translation invariant: if we add $30$ to the curve $y=f(x)$ and the curve $y=0$, the area of the region between them does not change. $\endgroup$ – André Nicolas Aug 15 '13 at 15:41
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You should notice that the function is negetive from 2 to $\infty$(because 2 is a max point and f decrease for every $x>2$), so you need to multi

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