KCET · Maths · Differentiation
Area bounded by \( y=x^{3}, y=8 \) and \( x=0 \) is
- A \( 2 \) sq. units
- B \( 14 \mathrm{sq} \). units
- C \( 12 \mathrm{sq} \). units
- D \( 6 \) sq. units
Answer & Solution
Correct Answer
(C) \( 12 \mathrm{sq} \). units
Step-by-step Solution
Detailed explanation
Given equation, \(y=x^{3} \rightarrow(1)\) \(y=8 \rightarrow(2)\) \(x=0 \rightarrow(3)\)
Now, \( x=y^{1 / 3} \) So, required area is given by \( \int_{0}^{8} x d y \) \( \begin{array}{l} =\int_{0}^{8} y^{1 / 3} d y=\left[\frac{3}{4} y^{4 / 3}\right]_{0}^{8}=\frac{3}{4}\left(8^{4 / 3}\right) \\ =\frac{3}{4}\left(2^{4}\right)=12 \end{array} \)
Now, \( x=y^{1 / 3} \) So, required area is given by \( \int_{0}^{8} x d y \) \( \begin{array}{l} =\int_{0}^{8} y^{1 / 3} d y=\left[\frac{3}{4} y^{4 / 3}\right]_{0}^{8}=\frac{3}{4}\left(8^{4 / 3}\right) \\ =\frac{3}{4}\left(2^{4}\right)=12 \end{array} \)See the Complete Solution
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