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Let $f(x,y)$ be a positive function. If the integrals

(A) $\int_{0}^{1} \int_{x^2}^{1} f(x,y) dydx$

(B) $\int_{0}^{1} \int_{x^3}^{1} f(x,y) dydx$

(C) $\int_{0}^{1} \int_{0}^{1} f(x,y) dydx$

are ranked from smallest to largest, then

(a) (A) < (C) < (B)

(b) (B) < (A) < (C)

(c) (A) < (B) < (C)

(d) (C) < (A) < (B)

(e) (B) < (C) < (A)

(f) None of the above.

For this question, I got the answer (c), and I just want to verify that my work is correct:

I graphed the lines $y=x^2, y=x^3, x=0, x=1, y=0, y=1$

$y=x^2$ is green

$y=x^3$ is purple

So just by looking at this graph, I can tell that (C) is the largest, (B) is the second largest, and (A) is the smallest.

So the order should be (A)<(B)<(C)

Thus my answer is choice (c).

Is this correct?

enter image description here

Stephen Donovan
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user477465
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  • But I thought I am not looking at the region "under" $y=x^2$ and instead I'm looking for the region between $y=1$ and $y=x^2$ and bounded by $x=0$ and $x=1$ – user477465 May 25 '22 at 21:42
  • So the region between $y=1$ and $y=x^2$ and bounded by $x=0$ and $x=1$ has less area than that of $y=1$ and $y=x^3$ between $x=0$ and $x=1$? – user477465 May 25 '22 at 21:44
  • Deleted my comment; I think I made the mistake that $f$ is positive and constant, but that's clearly not the case. – user170231 May 26 '22 at 17:30
  • You're right. I didn't read carefully enough. The title is incorrect. You you must use the fact that $f$ is positive; it's not just about areas. But (c) is correct. – Ted Shifrin May 27 '22 at 04:14

0 Answers0