Exam

NTA Abhyas

Year

2020

Subject

Mathematics

Chapter

Area Under Curves

NTA Abhyas 2020 · Mathematics · Area Under Curves

The area bounded by the curves y = x^ 2 and y = 2 (1 + x^ 2 ) is

The area bounded by the curves $y = x^{2}$ and $y = 2 \left(1 + x^{2}\right) is$

Step-by-step solution

For point of intersection $x^{2} = \dfrac{2}{1 + x^{2}}$

$⇒ x^{4} + x^{2} - 2 = 0$

$⇒ \left(x^{2} - 1\right)\left(x^{2} + 2\right) = 0$

$⇒ x = 1, - 1, ⇒ y = 1, -1$

$∴ area = 2\mathop{∫}_{0}^{1}\left(\dfrac{2}{1 + x^{2}} - x^{2}\right)dx$

$= 4{\left[\tan^{-1} x\right]}_{0}^{1} - \dfrac{2}{3}{\left[x^{3}\right]}_{0}^{1}$

$= 4\left(\pi / 4\right) - 2 / 3 = \pi - 2 / 3$