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A plane figure is bounded by the curves $2y=x^2$ and $x^3y=16$, the x-axis and the ordinate at x=4. Calculate the area enclosed.

My attempt:

$2y=x^2,\; x^3y=16 \\ y=\frac{x^2}{2} \quad y= \frac{16}{x^3}\\$

$\int ^?_?\biggr[\frac{x^2}{2} -\frac{16}{x^3} \biggr] \mathrm{dx}$

My question is what to use as the upper and lower limits.

I had solved similar a question: Find the area enclosed by the curve $a(a-x)y=x^3$, the x-axis and the line $2x=a$

Which I used $0$ and $x=\frac{a}{2}$ as the lower and the upper limits and I got the answer. I later became skeptical about my assumption in this question.

How do I really get to know the limits?

Orestes Dante
  • 549

2 Answers2

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Hint: We get from $$\frac{x^2}{2}=\frac{16}{x^3}$$ the equation $$x^5=32$$ And we have to integrate $$\int_{0}^{2}\frac{x^2}{2}dx+\int_{2}^{4}\frac{16}{x^3}dx$$

Dr. Sonnhard Graubner
  • 95,283
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Desmos is your friend.

$$\int_{0}^{2}\frac{x^{2}}{2}dx+\int_{2}^{4}16x^{-3}dx=\frac{4}{3}+\frac{3}{2}=\frac{17}{6}$$