Class 12 Mathematics Chapter 8 Application of Integrals

Key Concept: Area Calculation

c) $$\frac{15}{4}$$
[Solution Description]
The area under the curve $$y = x^3$$ from $$x = 1$$ to $$x = 2$$ is given by the integral $$\int_1^2 x^3 \, dx$$. Evaluating this integral, we have:
$$
\int_1^2 x^3 \, dx = \left[\frac{x^4}{4}\right]_1^2 = \frac{2^4}{4} – \frac{1^4}{4} = \frac{16}{4} – \frac{1}{4} = \frac{15}{4}.
$$
Therefore, the area is $$\frac{15}{4}$$.

Key Concept: Application of Formula

c) 24

[Solution Description] To find the area under the line $$y = 3x$$, we use the integral from 0 to 4:

$$\int_{0}^{4} 3x \, dx = \left[\frac{3x^2}{2}\right]_{0}^{4} = \frac{3(4)^2}{2} – \frac{3(0)^2}{2} = \frac{48}{2} = 24.$$

Thus, the area is 24 square units.

Key Concept: Rectangle Area Calculation

c) 21
[Solution Description] The area of a rectangle is given by $$A = \text{length} \times \text{width}$$. Thus, $$A = 7 \times 3 = 21$$ square units.

Key Concept: Conceptual Understanding

d) $$ dy $$
[Solution Description] In the context of horizontal strips, the integration is performed with respect to the variable $$ y $$, as the function is expressed in terms of $$ y $$.

Key Concept: Composite Areas

d) $$\frac{32}{3}$$

[Solution Description]

The area enclosed by $$y = x^2$$ and $$y = 4$$ consists of two regions: one above and one below the parabola.

First, calculate the intersection points by setting $$x^2 = 4$$:

$$x = \pm 2.$$

Evaluate the area from $$x = -2$$ to $$x = 2$$:

$$A = \int_{-2}^{2} (4 – x^2) \, dx.$$

Find the antiderivative:

$$A = [4x – \frac{x^3}{3}]_{-2}^{2}.$$

Evaluate at the bounds:

$$A = \left[4(2) – \frac{8}{3}\right] – \left[4(-2) + \frac{-8}{3}\right].$$

Simplify:

$$A = \left(8 – \frac{8}{3}\right) – \left(-8 + \frac{8}{3}\right).$$

Further simplification yields:

$$A = (8 + 8) – \left(\frac{8}{3} + \frac{8}{3}\right),$$

$$A = 16 – \frac{16}{3},$$

$$A = \frac{48}{3} – \frac{16}{3} = \frac{32}{3}.$$

Hence, the area is $$\frac{32}{3}$$ square units.

Key Concept: Basic Integration

c) 24
[Solution Description] Compute the definite integral:

$$
\int_{0}^{4} 3x \, dx = \left[\frac{3x^2}{2}\right]_{0}^{4}.
$$

Evaluate this:

$$
= \frac{3 \times 4^2}{2} – \frac{3 \times 0^2}{2}
= \frac{48}{2} – 0
= 24.
$$

Therefore, the area under the curve is 24 square units.

Key Concept: Quadrant Area Calculation

c) $$\frac{25\pi}{4}$$
[Solution Description]
The equation of the circle $$x^2 + y^2 = 25$$ indicates that the radius $$ r = 5 $$. The total area of the circle is:
$$
A = \pi \times 5^2 = 25\pi.
$$
The area of the circle in the first quadrant (one-fourth of the circle) is:
$$
\frac{25\pi}{4}.
$$
Therefore, the area is $$ \frac{25\pi}{4} $$ which simplifies to approximately $$6.25\pi$$.

Key Concept: Absolute Value of Area

c) Take the absolute value
[Solution Description] When the computed area comes out as negative because the curve lies below the x-axis, the absolute value of that negative area should be taken to represent the physical area correctly. Thus, the correct solution is to take the absolute value.

Key Concept: First Quadrant Area

a) $$4\pi$$ square units
[Solution Description] Since the radius of the circle is 4, the area of the first quadrant part is a quarter of the total circle’s area:
$$
A = \frac{1}{4} \times \pi \times r^2 = \frac{1}{4} \times \pi \times 4^2 = \frac{1}{4} \times 16\pi = 4\pi
$$
Therefore, the area is $$4\pi$$ square units.

Key Concept: Definite Integral Evaluation

b) 8
[Solution Description]
To find the area under this curve, evaluate the integral:
$$
\int_1^3 2x \, dx
$$
Calculating the integral:
\begin{align*}
\int 2x \, dx &= [x^2]_1^3 \\
&= (3^2) – (1^2) \\
&= 9 – 1 \\
&= 8
\end{align*}

Key Concept: Simple Substitution

c) $$18\pi$$
[Solution Description] Using the formula for the area of an ellipse $$A = \pi ab$$, substitute $$a = 6$$ and $$b = 3$$:

$$
A = \pi \times 6 \times 3 = 18\pi.
$$

Therefore, the area of the ellipse is $$18\pi$$.

Key Concept: Horizontal Strip Method

b) 9
[Solution Description] The given boundaries form a region where $$x = y^2$$ and $$y$$ varies from 0 to 3. Using horizontal strips, integrate with respect to $$y$$. The area is given by:
$$
A = \int_{0}^{3} y^2 \, dy
$$
The antiderivative of $$ y^2 $$ is $$ \frac{y^3}{3} $$:
$$
A = \left[\frac{y^3}{3}\right]_{0}^{3} = \frac{27}{3} – \frac{0}{3} = 9
$$

Key Concept: Complex Curves, Real-world Application

b) 16 cubic meters
[Solution Description]
The volume $$V$$ of the water in the tank can be found by integrating the cross-sectional area along its length. The area $$A$$ under the curve $$y = \sqrt{x}$$ from $$x = 0$$ to $$x = 4$$ is given by:

$$
A = \int_{0}^{4} \sqrt{x} \, dx
$$

Solving this integral, we have:

$$
A = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{4} = \frac{2}{3} \times (4^{3/2}) – \frac{2}{3} \times (0^{3/2})
$$

Calculating $$4^{3/2}$$ gives $$8$$, so:

$$
A = \frac{2}{3} \times 8 = \frac{16}{3} \text{ square meters}
$$

Therefore, the total volume $$V$$ is:

$$
V = A \times \text{length} = \frac{16}{3} \times 3 = 16 \text{ cubic meters}
$$

Key Concept: Simple Function Identification

b) $$y = \sin x$$
[Solution Description] Sinusoidal curves are represented by functions like $$y = \sin x$$ or $$y = \cos x$$. Among the given options, only option b) $$y = \sin x$$ fits this description.

Key Concept: Area Formula

a) $$\int_{a}^{b} f(x) \, dx$$
[Solution Description]
The area of the region bounded by the curve, the x-axis, and vertical lines is given by integrating $$f(x)$$ with respect to $$x$$ from $$a$$ to $$b$$.
$$
Area = \int_{a}^{b} f(x) \, dx.
$$
This formula calculates the area between the curve and the x-axis.

Key Concept: Absolute Area

a) $$\frac{8}{3}$$ square units
[Solution Description] First, find the roots of $$y = x^2 – 2x = 0$$ to locate intersection points with the x-axis.

The roots are $$x(x – 2) = 0$$, thus $$x = 0$$ and $$x = 2$$.

We split the integral at $$x = 2$$ because the curve crosses the x-axis there:

$$
A_1 = \int_{0}^{2} |x^2 – 2x| \, dx = \int_{0}^{2} (2x – x^2) \, dx
$$

Evaluating it:

$$
A_1 = \left[ x^2 – \frac{x^3}{3} \right]_{0}^{2}
$$

Calculating,

$$
A_1 = \left( 2^2 – \frac{2^3}{3} \right) – 0 = \left( 4 – \frac{8}{3} \right) = \frac{12}{3} – \frac{8}{3} = \frac{4}{3}
$$

Next, compute the area from $$x = 2$$ to $$x = 3$$ as follows:

$$
A_2 = \int_{2}^{3} x^2 – 2x \, dx
$$

Solving:

$$
A_2 = \left[ \frac{x^3}{3} – x^2 \right]_{2}^{3}
$$

Calculate it,

$$
A_2 = \left( \frac{27}{3} – 9 \right) – \left( \frac{8}{3} – 4 \right) = \left( 9 – 9 \right) – \left( \frac{8}{3} – \frac{12}{3} \right) = 0 + \frac{4}{3}
$$

Total area $$A = A_1 + A_2 = \frac{8}{3}$$ square units.

Key Concept: Parabola Bounded Area

c) $$9$$
[Solution Description] The area can be calculated using vertical strips from $$y = 0$$ to $$y = 3$$.
$$
A = \int_{0}^{3} y^2 \, dy = \left[\frac{y^3}{3}\right]_{0}^{3} = \frac{27}{3} – 0 = 9
$$