CBSE 2025, Set 2

Question

The area of the shaded region (figure) represented by the curves $y=x^{2}$, $0\le x\le2$ and y-axis is given by :

![Diagram Area Under Curve](https://www.edvaya.com/images/cbse/2025/mathematics/set2/q7-area-under-curve.webp) A). $\int_{0}^{2}x^{2}dx$ B). $\int_{0}^{2}\sqrt{y}dy$ C). $\int_{0}^{4}x^{2}dx$ D). $\int_{0}^{4}\sqrt{y}dy$

Edvaya · Accepted Answer

The correct answer is $\int_{0}^{4}\sqrt{y}dy$. **Solution:**

The diagram shows the shaded region bounded by the curve $y = x^2$, the y-axis ($x = 0$), and the horizontal line $y = 4$.

When finding the area enclosed between a curve and the **y-axis**, we integrate with respect to $y$. The formula is:
$$	ext{Area} = \int_{y_1}^{y_2} x \,dy$$

From the given curve $y = x^2$, we can express $x$ in terms of $y$:
$$x = \sqrt{y}$$ (since we are in the first quadrant, $x \ge 0$).

The limits of integration along the y-axis are from the origin $y = 0$ to the top boundary $y = 4$ (which corresponds to $x = 2$ on the curve $y = x^2$).

Therefore, the area is given by:
$$\int_{0}^{4} \sqrt{y} \,dy$$

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Mathematics > Calculus > Application Of Integrals

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