KCET · Maths · Area Under Curves
The area of the region bounded by the line \(y=x\) and the curve \(y=x^3\) is
- A 0.2 sq unit
- B 0.3 sq unit
- C 0.4 sq unit
- D 0.5 sq unit
Answer & Solution
Correct Answer
(D) 0.5 sq unit
Step-by-step Solution
Detailed explanation
\(\because y=x\) and \(y=x\)
The curve intersect at \(x=-1, y_i=-1 ; x=0, y=0\) and \(x=1, y=1\).
Therefore, area \(=-\int_{-1}^0\left(x^3-x\right) d x+\int_0^1\left(x-x^3\right) d x\)
\(=-\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_{-1}^0+\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1\)
\(=-\left(\frac{1}{4}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)\)
\(=-\frac{1}{2}+1=\frac{1}{2}=0.5\) sq. units
The curve intersect at \(x=-1, y_i=-1 ; x=0, y=0\) and \(x=1, y=1\).
Therefore, area \(=-\int_{-1}^0\left(x^3-x\right) d x+\int_0^1\left(x-x^3\right) d x\)
\(=-\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_{-1}^0+\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1\)
\(=-\left(\frac{1}{4}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)\)
\(=-\frac{1}{2}+1=\frac{1}{2}=0.5\) sq. units
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