Find the area of the figure bounded by the given pair of curves y = x2 – x + 3 and y = 3
- 17/6 units (sq)
- 7/6 units (sq)
- 5/6 units (sq)
-
1/6 units (sq)
The correct answer is: D
Explanation
Area bounded by
y = \(x^2\) - x + 3 and y = 3
The intersection is given as \(x^2\) - x + 3 = 3
\(x^2\) - x = 0
x(x -1) = 0
The graph is sketched between x= 0 and x = 1
\(\int^{1}_{0}\)\(x^2\) - x + 3)dx - \(\int^{1}_{0}\)3dx = \(\int^{1}{0}(x^2\) - x)dx
[\(\frac{2x^3}{3} - \frac{x^2}{2}\)] = f(1) - f(0) = [\(\frac{2}{3} - \frac{1}{2}\)] - [0 - 0]
= \(\frac{4 - 3}{6}\) = \(\frac{1}{6}\)unit(sq)