The area of the region bounded by the curves | vedclass

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(B) The curves $y=x^{3}$ and $y=\frac{1}{x}$ intersect at $x^{3} = \frac{1}{x}$, which implies $x^{4} = 1$. Since we are in the first quadrant, $x=1$.

Vedclass · Accepted Answer

$\frac{1}{4}+\log _{e} 2$

Vedclass · Answer

(B) The curves $y=x^{3}$ and $y=\frac{1}{x}$ intersect at $x^{3} = \frac{1}{x}$, which implies $x^{4} = 1$. Since we are in the first quadrant, $x=1$. At $x=1$, $y=1$. The region is bounded by $y=x^{3}$ from $x=0$ to $x=1$ and by $y=\frac{1}{x}$ from $x=1$ to $x=2$. Required Area $= \int_{0}^{1} x^{3} dx + \int_{1}^{2} \frac{1}{x} dx$ $= \left[ \frac{x^{4}}{4} \right]_{0}^{1} + \left[ \log_{e} x \right]_{1}^{2}$ $= (\frac{1}{4} - 0) + (\log_{e} 2 - \log_{e} 1)$ $= \frac{1}{4} + \log_{e} 2$.

The area of the region bounded by the curves $y=x^{3}$, $y=\frac{1}{x}$, and the line $x=2$ in the first quadrant is:

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