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Q1(i):
Find the area under the given curves and given lines: (i) $y = x^2$, $x = 1$, $x = 2$ and $x$-axis

Solution :

Step 1: Identifying the Region and Theoretical Setup

The problem asks for the area of the region bounded by the parabola $y = x^2$, the vertical lines $x = 1$ and $x = 2$, and the $x$-axis. Since the function $y = x^2$ is continuous and non-negative for all $x \in [1, 2]$, the area $A$ is given by the definite integral of the function $f(x)$ with respect to $x$ over the interval $[a, b]$.

Formula: $A = \int_{a}^{b} |f(x)| \, dx = \int_{1}^{2} x^2 \, dx$ [Per the Fundamental Theorem of Calculus].

x y x=1 x=2 y=x^2

Step 2: Performing the Integration

Apply the Power Rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, where $n = 2$.

$$A = \int_{1}^{2} x^2 \, dx$$

$$A = \left[ \frac{x^{2+1}}{2+1} \right]_{1}^{2}$$

$$A = \left[ \frac{x^3}{3} \right]_{1}^{2}$$

Step 3: Applying Limits of Integration

Per the Second Fundamental Theorem of Calculus, substitute the upper limit and subtract the evaluation at the lower limit:

$$A = \left( \frac{2^3}{3} \right) - \left( \frac{1^3}{3} \right)$$

$$A = \frac{8}{3} - \frac{1}{3}$$

$$A = \frac{7}{3}$$

Summary of Calculation

Component Value
Function $f(x)$ $x^2$
Lower Bound ($a$) $1$
Upper Bound ($b$) $2$
Resulting Area $7/3$ square units

Final Solution: The area under the curve is $\frac{7}{3}$ square units.



CBSE Solutions for Class 12 Mathematics Application of Integrals


Chapters in CBSE - Class 12 Mathematics


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