Find the area of the region (in sq. units) bounded by the curve y = x, y = x^2 + 2 for x (-1, 1) . - Mathematics MCQ Mathematics Mock Test - 2 2026 | ExamTest.live

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Find the area of the region (in sq. units) bounded by the curve y = x, y = x^{2} + 2 for x \in (- 1, 1) .

Solution & Step-by-step Explanation

The area between two curves y_{1} and y_{2} from x = a to x = b is: A re a = \int_{a}^{b} ∣ y_{2} - y_{1} ∣ d x In the interval (- 1, 1), x^{2} + 2 is always greater than x . A re a = \int_{- 1}^{1} (x^{2} + 2 - x) d x A re a = [\frac{x ^{3}}{3} + 2 x - \frac{x ^{2}}{2}]_{- 1}^{1} At x = 1 : \frac{1}{3} + 2 - \frac{1}{2} = \frac{2 + 12 - 3}{6} = \frac{11}{6} At x = - 1 : \frac{- 1}{3} - 2 - \frac{1}{2} = \frac{- 2 - 12 - 3}{6} = - \frac{17}{6} $ A re a = \frac{11}{6} - (- \frac{17}{6}) = \frac{11 + 17}{6} = \frac{28}{6} = \frac{14}{3} $

Try it yourself before checking the explanation above.

Find the area of the region (in sq. units) bounded by the curve y = x, y = x^{2} + 2 for x \in (- 1, 1) .

11/3

1/3

2/3

14/3

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