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Hi there's one problem on my study guide that my teacher didn't go over and I don't know how to approach/solve it. Here's the problem:

Find the area between the curves on the given interval. Draw a graph of the functions and the region.

$$y=x^4 , y=x-1, -2 \le x \le 0.$$

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Hint: determine if there are any intersection points of these two curves on the specified intervals. If there are none, great! Determine which one of the two is greater on the interval (WLOG call the greater of the two $y_{1}$, and the smaller $y_{2}$; graphing should help you determine this!), and then find $$\int_{-2}^{0}y_{1}-y_{2} dx$$ If there are intersection points, break up the integral into several subintervals of $[0, 2]$ and again determine which curve is greater on each portion of the integral. Then integrate each piece exactly as you did for the case that there are no intersection points.

Edit: at intersection points, the function which is greater might change, and so you would have to break up the integral accordingly to get the right subtraction of functions. In this case, $x^{4}$ and $x-1$ don't intersect each other on the interval $[-2, 0]$, and so you don't have to worry about it; $x^{4} > x - 1$ for all $-2 \leq x \leq 0$.

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  • $\begingroup$ Ah much simpler than I thought. Thanks again! Edit: If there were intersection points, what numbers would I take the integrals of? Can you be more specific with your second hint up there? $\endgroup$ – user79477 May 27 '13 at 2:32
  • $\begingroup$ @user79477, I've edited the post accordingly. $\endgroup$ – Alex Wertheim May 27 '13 at 2:42
  • $\begingroup$ Thank you. So if I had a problem where I solved for intersection points that were, let's say x=1 and x=-5, I would use -5 and 1 as my integral range? What happens if you have 3 intersection points? $\endgroup$ – user79477 May 27 '13 at 2:46
  • $\begingroup$ It depends on whether or not your areas are restricted to a specified subinterval; in the case above, you were given a specific interval, and then had to consider intersections over that interval separately. Intersection points naturally define areas between two curves, and so if no interval is specified, then the intersection points are the natural interval. If there are multiple intersection points, you must partition the integral into several integrals, with bounds at each of the intersection points, taking into account which function is greater over each interval. $\endgroup$ – Alex Wertheim May 27 '13 at 2:56
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Let $E$ the region given by two curves $f$, $g$, which are functions with domain $[a,b]$ in the $x$ axis and $f>g$. Then:

$$A(E)=\iint_E 1\, dA=\int_a^b\int_{g(x)}^{f(x)} 1\, dydx=\int_a^b (f(x)-g(x))\,dx=\int_a^b f-\int_a^b g $$

Separe in cases when $f>g$ and $f\leq g$.

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  • $\begingroup$ Great. I got it now. Thank you. $\endgroup$ – user79477 May 27 '13 at 2:34

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