CBSE Class 12 Mathematics: Applications of Integrals NCERT Solutions

Solution:

The given curve is $y^2\; =\; x$. This is a parabola opening to the right. We need to find the area of the region bounded by this curve, the vertical lines $x\; =\; 1$ and $x\; =\; 4$, and the x-axis. Since the curve is symmetric about the x-axis and we are considering the area bounded by the x-axis, we will consider the area in the first quadrant and then double it, or simply integrate $y\; =\; \backslash sqrt\{x\}$ from $x=1$ to $x=4$.

From $y^2\; =\; x$, we get $y\; =\; \backslash pm\; \backslash sqrt\{x\}$. For the area in the first quadrant, we take $y\; =\; \backslash sqrt\{x\}$.

The area A is given by the definite integral:

$$A\; =\; \backslash int\_\{1\}^\{4\}\; y\; \backslash ,\; dx$$

Substitute $y\; =\; \backslash sqrt\{x\}$:

$$A\; =\; \backslash int\_\{1\}^\{4\}\; \backslash sqrt\{x\}\; \backslash ,\; dx$$

To evaluate this integral, we use the power rule for integration $\backslash int\; x^n\; dx\; =\; \backslash frac\{x^\{n+1\}\}\{n+1\}$. Here, $n\; =\; \backslash frac\{1\}\{2\}$.

$$A\; =\; \backslash left[\backslash frac\{x^\{\backslash frac\{1\}\{2\}+1\}\}\{\backslash frac\{1\}\{2\}+1\}\backslash right]\_\{1\}^\{4\}\; =\; \backslash left[\backslash frac\{x^\{\backslash frac\{3\}\{2\}\}\}\{\backslash frac\{3\}\{2\}\}\backslash right]\_\{1\}^\{4\}\; =\; \backslash left[\backslash frac\{2\}\{3\}\; x^\{\backslash frac\{3\}\{2\}\}\backslash right]\_\{1\}^\{4\}$$

Now, we apply the limits of integration:

$$A\; =\; \backslash frac\{2\}\{3\}\; \backslash left[\; (4)^\{\backslash frac\{3\}\{2\}\}\; -\; (1)^\{\backslash frac\{3\}\{2\}\}\; \backslash right]$$

Calculate the powers:

$(4)^\{\backslash frac\{3\}\{2\}\}\; =\; (\backslash sqrt\{4\})^3\; =\; 2^3\; =\; 8$

$(1)^\{\backslash frac\{3\}\{2\}\}\; =\; (\backslash sqrt\{1\})^3\; =\; 1^3\; =\; 1$

Substitute these values back:

$$A\; =\; \backslash frac\{2\}\{3\}\; [8\; -\; 1]\; =\; \backslash frac\{2\}\{3\}\; [7]\; =\; \backslash frac\{14\}\{3\}$$

The area of the region is $\backslash frac\{14\}\{3\}$ square units.

Solution:

The given curve is $y^2\; =\; 9x$. We need to find the area of the region bounded by this curve, the vertical lines $x\; =\; 2$ and $x\; =\; 4$, and the x-axis, specifically in the first quadrant.

From $y^2\; =\; 9x$, we get $y\; =\; \backslash pm\; \backslash sqrt\{9x\}\; =\; \backslash pm\; 3\backslash sqrt\{x\}$. Since we are considering the first quadrant, we take the positive root: $y\; =\; 3\backslash sqrt\{x\}$.

The area A is given by the definite integral:

$$A\; =\; \backslash int\_\{2\}^\{4\}\; y\; \backslash ,\; dx$$

Substitute $y\; =\; 3\backslash sqrt\{x\}$:

$$A\; =\; \backslash int\_\{2\}^\{4\}\; 3\backslash sqrt\{x\}\; \backslash ,\; dx$$

We can take the constant 3 outside the integral:

$$A\; =\; 3\; \backslash int\_\{2\}^\{4\}\; \backslash sqrt\{x\}\; \backslash ,\; dx$$

Using the power rule for integration $\backslash int\; x^n\; dx\; =\; \backslash frac\{x^\{n+1\}\}\{n+1\}$, with $n\; =\; \backslash frac\{1\}\{2\}$, the integral of $\backslash sqrt\{x\}$ is $\backslash frac\{2\}\{3\}\; x^\{\backslash frac\{3\}\{2\}\}$.

$$A\; =\; 3\; \backslash left[\backslash frac\{2\}\{3\}\; x^\{\backslash frac\{3\}\{2\}\}\backslash right]\_\{2\}^\{4\}$$

Simplify by canceling the 3s:

$$A\; =\; \backslash left[2\; x^\{\backslash frac\{3\}\{2\}\}\backslash right]\_\{2\}^\{4\}$$

Now, apply the limits of integration:

$$A\; =\; 2\; \backslash left[\; (4)^\{\backslash frac\{3\}\{2\}\}\; -\; (2)^\{\backslash frac\{3\}\{2\}\}\; \backslash right]$$

Calculate the powers:

$(4)^\{\backslash frac\{3\}\{2\}\}\; =\; (\backslash sqrt\{4\})^3\; =\; 2^3\; =\; 8$

$(2)^\{\backslash frac\{3\}\{2\}\}\; =\; 2\; \backslash sqrt\{2\}$

Substitute these values back:

$$A\; =\; 2\; [8\; -\; 2\backslash sqrt\{2\}]$$

Distribute the 2:

$$A\; =\; 16\; -\; 4\backslash sqrt\{2\}$$

The area of the region is $16\; -\; 4\backslash sqrt\{2\}$ square units.