CBSE Class 12 Maths Chapter 8: Application of Integrals NCERT Solutions

We are asked to find the area of the region bounded by the parabola $y^2\; =\; 9x$ and the straight line $y\; =\; 3x$. To find the area, we first need to determine the points of intersection of these two curves.

Given equations are:

1. $y^2\; =\; 9x$
2. $y\; =\; 3x$

Substitute the expression for $y$ from equation (2) into equation (1):

$$(3x)^2\; =\; 9x$$

$$9x^2\; =\; 9x$$

Rearrange the terms to form a quadratic equation:

$$9x^2\; -\; 9x\; =\; 0$$

Factor out $9x$:

$$9x(x\; -\; 1)\; =\; 0$$

This gives us two possible values for $x$: $x\; =\; 0$ or $x\; =\; 1$.

Now, find the corresponding $y$ values using $y\; =\; 3x$:

• If $x\; =\; 0$, then $y\; =\; 3(0)\; =\; 0$. So, one intersection point is (0, 0).
• If $x\; =\; 1$, then $y\; =\; 3(1)\; =\; 3$. So, the other intersection point is (1, 3).

The region is bounded by the parabola $y^2\; =\; 9x$ (which opens to the right) and the line $y\; =\; 3x$. For $x$ between 0 and 1, the line $y\; =\; 3x$ is above the curve $y\; =\; \backslash sqrt\{9x\}\; =\; 3\backslash sqrt\{x\}$ is not correct. Let's re-evaluate. For $x$ in (0, 1), say $x=1/4$, $y=3(1/4)=3/4$ for the line and $y=3\backslash sqrt\{1/4\}=3/2$ for the parabola. So the parabola is above the line. The area can be calculated by integrating the difference between the $y$ value of the parabola and the $y$ value of the line with respect to $x$, from $x=0$ to $x=1$.

The area $A$ is given by:

$$A\; =\; \backslash int\_\{0\}^\{1\}\; (\backslash text\{upper\; curve\}\; -\; \backslash text\{lower\; curve\})\; \backslash ,\; dx$$

From $y^2\; =\; 9x$, we get $y\; =\; \backslash pm\; 3\backslash sqrt\{x\}$. Since the line $y\; =\; 3x$ is in the first quadrant for $x\; >\; 0$, we consider the positive root $y\; =\; 3\backslash sqrt\{x\}$. Comparing $3\backslash sqrt\{x\}$ and $3x$ for $x\; \backslash in\; [0,\; 1]$, we find that $3\backslash sqrt\{x\}\; \backslash ge\; 3x$.

So, the area is:

$$A\; =\; \backslash int\_\{0\}^\{1\}\; (3\backslash sqrt\{x\}\; -\; 3x)\; \backslash ,\; dx$$

Factor out the constant 3:

$$A\; =\; 3\; \backslash int\_\{0\}^\{1\}\; (x^\{1/2\}\; -\; x)\; \backslash ,\; dx$$

Now, integrate term by term:

$$A\; =\; 3\; \backslash left[\; \backslash frac\{x^\{1/2\; +\; 1\}\}\{1/2\; +\; 1\}\; -\; \backslash frac\{x^\{1+1\}\}\{1+1\}\; \backslash right]\_\{0\}^\{1\}$$

$$A\; =\; 3\; \backslash left[\; \backslash frac\{x^\{3/2\}\}\{3/2\}\; -\; \backslash frac\{x^2\}\{2\}\; \backslash right]\_\{0\}^\{1\}$$

$$A\; =\; 3\; \backslash left[\; \backslash frac\{2\}\{3\}x^\{3/2\}\; -\; \backslash frac\{1\}\{2\}x^2\; \backslash right]\_\{0\}^\{1\}$$

Evaluate the expression at the limits:

$$A\; =\; 3\; \backslash left[\; \backslash left(\; \backslash frac\{2\}\{3\}(1)^\{3/2\}\; -\; \backslash frac\{1\}\{2\}(1)^2\; \backslash right)\; -\; \backslash left(\; \backslash frac\{2\}\{3\}(0)^\{3/2\}\; -\; \backslash frac\{1\}\{2\}(0)^2\; \backslash right)\; \backslash right]$$

$$A\; =\; 3\; \backslash left[\; \backslash left(\; \backslash frac\{2\}\{3\}\; -\; \backslash frac\{1\}\{2\}\; \backslash right)\; -\; (0\; -\; 0)\; \backslash right]$$

Find a common denominator for the fractions inside the bracket:

$$A\; =\; 3\; \backslash left[\; \backslash frac\{4\}\{6\}\; -\; \backslash frac\{3\}\{6\}\; \backslash right]$$

$$A\; =\; 3\; \backslash left[\; \backslash frac\{1\}\{6\}\; \backslash right]$$

$$A\; =\; \backslash frac\{3\}\{6\}\; =\; \backslash frac\{1\}\{2\}$$

The area of the region bounded by the curves is $1/2$ square units.

We need to find the area enclosed between two parabolas: $y^2\; =\; 2px$ and $x^2\; =\; 2py$. First, let's find their points of intersection.

From the second equation, $x^2\; =\; 2py$, we can express $y$ in terms of $x$: $y\; =\; \backslash frac\{x^2\}\{2p\}$.

Substitute this expression for $y$ into the first equation, $y^2\; =\; 2px$:

$$\backslash left(\; \backslash frac\{x^2\}\{2p\}\; \backslash right)^2\; =\; 2px$$

$$\backslash frac\{x^4\}\{4p^2\}\; =\; 2px$$

Multiply both sides by $4p^2$:

$$x^4\; =\; 8p^3x$$

Rearrange the terms to solve for $x$:

$$x^4\; -\; 8p^3x\; =\; 0$$

Factor out $x$:

$$x(x^3\; -\; 8p^3)\; =\; 0$$

This equation gives us two possible solutions for $x$:

• $x\; =\; 0$
• $x^3\; -\; 8p^3\; =\; 0\; \backslash implies\; x^3\; =\; 8p^3\; \backslash implies\; x\; =\; 2p$

Now, find the corresponding $y$ values using $y\; =\; \backslash frac\{x^2\}\{2p\}$:

• If $x\; =\; 0$, then $y\; =\; \backslash frac\{0^2\}\{2p\}\; =\; 0$. The intersection point is (0, 0).
• If $x\; =\; 2p$, then $y\; =\; \backslash frac\{(2p)^2\}\{2p\}\; =\; \backslash frac\{4p^2\}\{2p\}\; =\; 2p$. The intersection point is (2p, 2p).

The region is bounded by the two parabolas between $x\; =\; 0$ and $x\; =\; 2p$. To find the area, we need to determine which curve is above the other in this interval. The first parabola is $y\; =\; \backslash sqrt\{2px\}$ (considering the upper half) and the second is $y\; =\; \backslash frac\{x^2\}\{2p\}$.

Let's test a value between 0 and $2p$, for example, $x\; =\; p$:

• For $y^2\; =\; 2px$, $y\; =\; \backslash sqrt\{2p(p)\}\; =\; \backslash sqrt\{2p^2\}\; =\; p\backslash sqrt\{2\}$.
• For $x^2\; =\; 2py$, $y\; =\; \backslash frac\{p^2\}\{2p\}\; =\; \backslash frac\{p\}\{2\}$.

Since $p\backslash sqrt\{2\}\; \backslash approx\; 1.414p$ and $p/2\; =\; 0.5p$, we see that $\backslash sqrt\{2px\}\; \backslash ge\; \backslash frac\{x^2\}\{2p\}$ for $x\; \backslash in\; [0,\; 2p]$. Therefore, $y\; =\; \backslash sqrt\{2px\}$ is the upper curve and $y\; =\; \backslash frac\{x^2\}\{2p\}$ is the lower curve.

The area $A$ is given by the integral of the difference between the upper and lower curves:

$$A\; =\; \backslash int\_\{0\}^\{2p\}\; \backslash left(\; \backslash sqrt\{2px\}\; -\; \backslash frac\{x^2\}\{2p\}\; \backslash right)\; \backslash ,\; dx$$

Separate the terms and integrate:

$$A\; =\; \backslash int\_\{0\}^\{2p\}\; \backslash sqrt\{2p\}\; x^\{1/2\}\; \backslash ,\; dx\; -\; \backslash int\_\{0\}^\{2p\}\; \backslash frac\{1\}\{2p\}\; x^2\; \backslash ,\; dx$$

$$A\; =\; \backslash sqrt\{2p\}\; \backslash left[\; \backslash frac\{x^\{3/2\}\}\{3/2\}\; \backslash right]\_\{0\}^\{2p\}\; -\; \backslash frac\{1\}\{2p\}\; \backslash left[\; \backslash frac\{x^3\}\{3\}\; \backslash right]\_\{0\}^\{2p\}$$

$$A\; =\; \backslash sqrt\{2p\}\; \backslash left[\; \backslash frac\{2\}\{3\}x^\{3/2\}\; \backslash right]\_\{0\}^\{2p\}\; -\; \backslash frac\{1\}\{2p\}\; \backslash left[\; \backslash frac\{x^3\}\{3\}\; \backslash right]\_\{0\}^\{2p\}$$

Evaluate at the limits:

$$A\; =\; \backslash sqrt\{2p\}\; \backslash left(\; \backslash frac\{2\}\{3\}(2p)^\{3/2\}\; -\; 0\; \backslash right)\; -\; \backslash frac\{1\}\{2p\}\; \backslash left(\; \backslash frac\{(2p)^3\}\{3\}\; -\; 0\; \backslash right)$$

$$A\; =\; \backslash sqrt\{2p\}\; \backslash left(\; \backslash frac\{2\}\{3\}\; (2\backslash sqrt\{2\}\; p^\{3/2\})\; \backslash right)\; -\; \backslash frac\{1\}\{2p\}\; \backslash left(\; \backslash frac\{8p^3\}\{3\}\; \backslash right)$$

$$A\; =\; \backslash frac\{4\backslash sqrt\{2\}\}\{3\}\; p^\{1/2\}\; p^\{3/2\}\; -\; \backslash frac\{8p^3\}\{6p\}$$

$$A\; =\; \backslash frac\{4\backslash sqrt\{2\}\}\{3\}\; p^2\; -\; \backslash frac\{4\}\{3\}\; p^2$$

Combine the terms:

$$A\; =\; \backslash frac\{4p^2\}\{3\}\; (\backslash sqrt\{2\}\; -\; 1)$$

Let's recheck the calculation. The source calculation seems to have a different result. Let's re-evaluate the source calculation:

$$=\; \backslash sqrt\{2p\}\; \backslash left[\; \backslash frac\{2(x)^\{3/2\}\}\{3\}\; \backslash right]\_\{0\}^\{2p\}\; -\; \backslash frac\{1\}\{2p\}\; \backslash left[\; \backslash frac\{x^3\}\{3\}\; \backslash right]\_\{0\}^\{2p\}$$

$$=\; \backslash sqrt\{2p\}\; \backslash left[\; \backslash frac\{2\}\{3\}\; \backslash cdot\; (2p)^\{3/2\}\; -\; 0\; \backslash right]\; -\; \backslash frac\{1\}\{2p\}\; \backslash left[\; \backslash frac\{1\}\{3\}\; (2p)^3\; -\; 0\; \backslash right]$$

$$=\; \backslash sqrt\{2p\}\; \backslash left(\; \backslash frac\{2\}\{3\}\; \backslash cdot\; 2\backslash sqrt\{2\}\; p^\{3/2\}\; \backslash right)\; -\; \backslash frac\{1\}\{2p\}\; \backslash left(\; \backslash frac\{8\; p^3\}\{3\}\; \backslash right)$$

$$=\; \backslash frac\{4\backslash sqrt\{2\}\}\{3\}\; p^\{1/2\}\; p^\{3/2\}\; -\; \backslash frac\{8p^3\}\{6p\}$$

$$=\; \backslash frac\{4\backslash sqrt\{2\}\}\{3\}\; p^2\; -\; \backslash frac\{4\}\{3\}\; p^2$$

$$=\; \backslash frac\{4p^2\}\{3\}\; (\backslash sqrt\{2\}\; -\; 1)$$

The source calculation has:

$$=\; \backslash sqrt\{2p\}\; \backslash left(\; \backslash frac\{2\}\{3\}\; \backslash cdot\; 2\backslash sqrt\{2\}\; p^\{3/2\}\; \backslash right)\; -\; \backslash frac\{1\}\{2p\}\; \backslash left(\; \backslash frac\{1\}\{3\}\; 8\; p^3\; \backslash right)$$

$$=\; \backslash sqrt\{2\}\; p\; \backslash left(\; \backslash frac\{4\backslash sqrt\{2\}\}\{3\}\; p^\{3/2\}\; \backslash right)\; -\; \backslash frac\{1\}\{2p\}\; \backslash left(\; \backslash frac\{8\}\{3\}\; p^3\; \backslash right)$$

There seems to be a mistake in copying $\backslash sqrt\{2p\}$ as $\backslash sqrt\{2\}p$ in the source's intermediate step. Let's correct it.

$$A\; =\; \backslash sqrt\{2p\}\; \backslash left(\; \backslash frac\{2\}\{3\}\; (2p)^\{3/2\}\; \backslash right)\; -\; \backslash frac\{1\}\{2p\}\; \backslash left(\; \backslash frac\{8p^3\}\{3\}\; \backslash right)$$

$$A\; =\; \backslash sqrt\{2p\}\; \backslash left(\; \backslash frac\{2\}\{3\}\; \backslash cdot\; 2\backslash sqrt\{2\}\; p\backslash sqrt\{p\}\; \backslash right)\; -\; \backslash frac\{4p^2\}\{3\}$$

$$A\; =\; \backslash frac\{4\backslash sqrt\{2\}\}\{3\}\; p\; \backslash sqrt\{2p\}\; -\; \backslash frac\{4p^2\}\{3\}$$

$$A\; =\; \backslash frac\{4\backslash sqrt\{2\}\}\{3\}\; p\; \backslash cdot\; \backslash sqrt\{2\}\; \backslash sqrt\{p\}\; -\; \backslash frac\{4p^2\}\{3\}$$

$$A\; =\; \backslash frac\{4\; \backslash cdot\; 2\}\{3\}\; p\; \backslash sqrt\{p\}\; -\; \backslash frac\{4p^2\}\{3\}$$

This is still not matching the source's final answer. Let's re-evaluate the source's calculation carefully.

Source calculation:

$$=\; \backslash sqrt\{2p\}\; \backslash left[\; \backslash frac\{2\}\{3\}\; \backslash cdot\; (2p)^\{3/2\}\; -\; 0\; \backslash right]\; -\; \backslash frac\{1\}\{2p\}\; \backslash left[\; \backslash frac\{1\}\{3\}\; (2p)^3\; -\; 0\; \backslash right]$$

$$=\; \backslash sqrt\{2p\}\; \backslash left(\; \backslash frac\{2\}\{3\}\; \backslash cdot\; 2\backslash sqrt\{2\}\; p^\{3/2\}\; \backslash right)\; -\; \backslash frac\{1\}\{2p\}\; \backslash left(\; \backslash frac\{8\; p^3\}\{3\}\; \backslash right)$$

Here, $(2p)^\{3/2\}\; =\; (2p)\backslash sqrt\{2p\}\; =\; 2p\backslash sqrt\{2\}\backslash sqrt\{p\}$. So,

$$=\; \backslash sqrt\{2p\}\; \backslash left(\; \backslash frac\{2\}\{3\}\; \backslash cdot\; 2\backslash sqrt\{2\}\; p\backslash sqrt\{p\}\; \backslash right)\; -\; \backslash frac\{4p^2\}\{3\}$$

$$=\; \backslash frac\{4\backslash sqrt\{2\}\}\{3\}\; p\; \backslash sqrt\{2p\}\backslash sqrt\{p\}\; -\; \backslash frac\{4p^2\}\{3\}$$

$$=\; \backslash frac\{4\backslash sqrt\{2\}\}\{3\}\; p\; \backslash cdot\; \backslash sqrt\{2\}\; \backslash cdot\; p^\{1/2\}\; \backslash cdot\; p^\{1/2\}\; -\; \backslash frac\{4p^2\}\{3\}$$

$$=\; \backslash frac\{4\; \backslash cdot\; 2\}\{3\}\; p^2\; -\; \backslash frac\{4p^2\}\{3\}\; =\; \backslash frac\{8p^2\}\{3\}\; -\; \backslash frac\{4p^2\}\{3\}\; =\; \backslash frac\{4p^2\}\{3\}$$

The source calculation seems to have an error in the simplification of $\backslash sqrt\{2p\}\; \backslash cdot\; p^\{3/2\}$. Let's use the correct values:

$$A\; =\; \backslash sqrt\{2p\}\; \backslash left(\; \backslash frac\{2\}\{3\}\; (2p)^\{3/2\}\; \backslash right)\; -\; \backslash frac\{1\}\{2p\}\; \backslash left(\; \backslash frac\{8p^3\}\{3\}\; \backslash right)$$

$$A\; =\; \backslash sqrt\{2p\}\; \backslash left(\; \backslash frac\{2\}\{3\}\; \backslash cdot\; 2p\; \backslash sqrt\{2p\}\; \backslash right)\; -\; \backslash frac\{4p^2\}\{3\}$$

$$A\; =\; \backslash frac\{4p\}\{3\}\; (\backslash sqrt\{2p\})^2\; -\; \backslash frac\{4p^2\}\{3\}$$

$$A\; =\; \backslash frac\{4p\}\{3\}\; (2p)\; -\; \backslash frac\{4p^2\}\{3\}$$

$$A\; =\; \backslash frac\{8p^2\}\{3\}\; -\; \backslash frac\{4p^2\}\{3\}$$

$$A\; =\; \backslash frac\{4p^2\}\{3\}$$

The area of the region bounded by the two parabolas is $\backslash frac\{4p^2\}\{3\}$ square units.