The area bounded by the curves $y = \cos x$ and $y = \sin x$ between the ordinates $x = 0$ and $x = \frac{3\pi}{2}$ is Options: (A) $4\sqrt{2}+2$ (B) $4\sqrt{2}-1$ (C) $4\sqrt{2}+1$ (D) $4\sqrt{2}-2$

### The Geometric Dance of Sine and Cosine Welcome, future engineer. Today, we are not just solving an integral; we are witnessing a geometric dance. Imagine the sine and cosine waves as two dancers on a stage, moving in perfect harmony, yet constantly crossing paths. Our goal is to measure the area of the stage they enclose between $x = 0$ and $x = \frac{3\pi}{2}$. This is not a simple calculation; it is a test of your ability to visualize the behavior of functions. ### Phase 1: Finding the Intersection First, we must find where these dancers cross. By setting $\sin x = \cos x$, we find the intersection points at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$. These points act as our boundaries, slicing the total area into three distinct regions. Without these boundaries, we would be blind to the changing hierarchy of the curves. ### Phase 2: The Calculus of Three Regions In the first region, from $0$ to $\frac{\pi}{4}$, the cosine curve is the leader, sitting above the sine curve. We integrate $(\cos x - \sin x)$ to find the area: $$A_1 = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) dx = \sqrt{2} - 1$$ As we move to the second region, from $\frac{\pi}{4}$ to $\frac{5\pi}{4}$, the sine curve takes the lead. We integrate $(\sin x - \cos x)$ to find the area: $$A_2 = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin x - \cos x) dx = 2\sqrt{2}$$ Finally, in the third region, from $\frac{5\pi}{4}$ to $\frac{3\pi}{2}$, the cosine curve returns to the top. We calculate the area as: $$A_3 = \int_{\frac{5\pi}{4}}^{\frac{3\pi}{2}} (\cos x - \sin x) dx = \sqrt{2} - 1$$ ### Phase 3: The Grand Finale By summing these individual regions, we arrive at the total area: $$A_{total} = A_1 + A_2 + A_3 = (\sqrt{2} - 1) + 2\sqrt{2} + (\sqrt{2} - 1)$$ The final result of this integration is: **$$A = 4\sqrt{2} - 2$$** Remember, the beauty of calculus lies in this systematic breakdown of complexity into simplicity. We did not just calculate an area; we mapped the relationship between two fundamental functions. Keep practicing, and you will master this.

JEE Main 2010

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: The area bounded by the curves $y = cos x$ and $y = sin x$ between the ordinates $x = 0$ and $x = \frac{3 π}{2}$ is

Q: How do I know which function is on top in each interval?

Pick a test point! For the interval $[0, \frac{\pi}{4}]$, choose $x = 0$. At $x=0$, $\cos(0) = 1$ and $\sin(0) = 0$. Since $1 > 0$, the cosine curve is on top. Do this for every interval to ensure your subtraction order is correct.

Q: Is there a symmetry I can use to save time?

Absolutely! Notice that the region from $0$ to $\frac{\pi}{4}$ and the region from $\frac{5\pi}{4}$ to $\frac{3\pi}{2}$ are symmetric. Calculating one and doubling it can save you precious minutes in the exam hall.

Select Answer:

Progressive Hints

Hint 1

Visualize the dance of the sine and cosine waves. They are not static; they cross each other. The area trapped between them is not a single, monolithic block, but a series of distinct pockets. Your first task is to identify where these curves intersect within the interval

[0, \frac{3 π}{2}]

Hint 2

Remember that the area between two curves

f (x)

and

g (x)

is defined by the integral of the absolute difference:

\int ∣ f (x) - g (x) ∣ d x

. Since the absolute value function is tricky to integrate directly, you must split the integral at the intersection points where the sign of

(f (x) - g (x))

changes.

Hint 3

Start your calculation by setting

sin x = cos x

to find the intersection points. Once you have your boundaries, set up the integral for the first region:

\int_{0}^{\frac{π}{4}} (cos x - sin x) d x

. Proceed similarly for the other two regions, ensuring you always subtract the lower curve from the upper curve.

Visualized Solution

The Curves and the Interval

We need to find the area bounded by $y = cos x$ and $y = sin x$ .
The given interval is $x \in [0, \frac{3 π}{2}]$ .

Identifying the Upper and Lower Curves

The area between two curves is given by $\int ∣ f (x) - g (x) ∣ d x$ .
We must determine which curve is on top in different intervals.
This requires finding their points of intersection.

Equating the Functions

To find intersection points, set the functions equal:
$sin x = cos x$
Dividing by $cos x$ , we get:
$tan x = 1$

Solving for Intersection Points

Solve $tan x = 1$ in the interval $[0, \frac{3 π}{2}]$ .
In the first quadrant: $x = \frac{π}{4}$ .
In the third quadrant: $x = π + \frac{π}{4} = \frac{5 π}{4}$ .

Splitting the Area

The total area is split into three distinct regions:
$A_{1}$ from $0$ to $\frac{π}{4}$
$A_{2}$ from $\frac{π}{4}$ to $\frac{5 π}{4}$
$A_{3}$ from $\frac{5 π}{4}$ to $\frac{3 π}{2}$

Setting up Integral for $A_{1}$

For $x \in [0, \frac{π}{4}]$ , the cosine curve is above the sine curve.
$A_{1} = \int_{0}^{\frac{π}{4}} (cos x - sin x) d x$

Calculating $A_{1}$

$A_{1} = [sin x - (- cos x)]_{0}^{\frac{π}{4}} = [sin x + cos x]_{0}^{\frac{π}{4}}$
$A_{1} = (sin \frac{π}{4} + cos \frac{π}{4}) - (sin 0 + cos 0)$
$A_{1} = (\frac{1}{2} + \frac{1}{2}) - (0 + 1) = 2 - 1$

Setting up Integral for $A_{2}$

For $x \in [\frac{π}{4}, \frac{5 π}{4}]$ , the sine curve is above the cosine curve.
$A_{2} = \int_{\frac{π}{4}}^{\frac{5 π}{4}} (sin x - cos x) d x$

Calculating $A_{2}$

$A_{2} = [- cos x - sin x]_{\frac{π}{4}}^{\frac{5 π}{4}}$
$A_{2} = (- cos \frac{5 π}{4} - sin \frac{5 π}{4}) - (- cos \frac{π}{4} - sin \frac{π}{4})$
$A_{2} = (\frac{1}{2} + \frac{1}{2}) - (- \frac{1}{2} - \frac{1}{2})$
$A_{2} = 2 - (- 2) = 22$

Setting up Integral for $A_{3}$

For $x \in [\frac{5 π}{4}, \frac{3 π}{2}]$ , the cosine curve is again above the sine curve.
$A_{3} = \int_{\frac{5 π}{4}}^{\frac{3 π}{2}} (cos x - sin x) d x$

Calculating $A_{3}$

$A_{3} = [sin x + cos x]_{\frac{5 π}{4}}^{\frac{3 π}{2}}$
$A_{3} = (sin \frac{3 π}{2} + cos \frac{3 π}{2}) - (sin \frac{5 π}{4} + cos \frac{5 π}{4})$
$A_{3} = (- 1 + 0) - (- \frac{1}{2} - \frac{1}{2})$
$A_{3} = - 1 - (- 2) = 2 - 1$

Total Area Calculation

$Total Area = A_{1} + A_{2} + A_{3}$
$Total Area = (2 - 1) + 22 + (2 - 1)$
$Total Area = 42 - 2$

00:00 / 00:00

The Sigma Insight: Area Bounded by Curves

Solution Diagram

The Geometric Dance of Sine and Cosine

Welcome, future engineer. Today, we are not just solving an integral; we are witnessing a geometric dance. Imagine the sine and cosine waves as two dancers on a stage, moving in perfect harmony, yet constantly crossing paths.

Our goal is to measure the area of the stage they enclose between

x = 0

and

x = \frac{3 π}{2}

. This is not a simple calculation; it is a test of your ability to visualize the behavior of functions.

Phase 1

Finding the Intersection

First, we must find where these dancers cross. By setting

sin x = cos x

, we find the intersection points at

x = \frac{π}{4}

and

x = \frac{5 π}{4}

These points act as our boundaries, slicing the total area into three distinct regions. Without these boundaries, we would be blind to the changing hierarchy of the curves.

Phase 2

The Calculus of Three Regions

In the first region, from

0

\frac{π}{4}

, the cosine curve is the leader, sitting above the sine curve. We integrate

(cos x - sin x)

to find the area:

A_{1} = \int_{0}^{\frac{π}{4}} (cos x - sin x) d x = 2 - 1

As we move to the second region, from

\frac{π}{4}

\frac{5 π}{4}

, the sine curve takes the lead. We integrate

(sin x - cos x)

to find the area:

A_{2} = \int_{\frac{π}{4}}^{\frac{5 π}{4}} (sin x - cos x) d x = 22

Finally, in the third region, from

\frac{5 π}{4}

\frac{3 π}{2}

, the cosine curve returns to the top. We calculate the area as:

A_{3} = \int_{\frac{5 π}{4}}^{\frac{3 π}{2}} (cos x - sin x) d x = 2 - 1

Phase 3

The Grand Finale

By summing these individual regions, we arrive at the total area:

A_{t o t a l} = A_{1} + A_{2} + A_{3} = (2 - 1) + 22 + (2 - 1)

The final result of this integration is:

$A = 42 - 2$

Remember, the beauty of calculus lies in this systematic breakdown of complexity into simplicity. We did not just calculate an area; we mapped the relationship between two fundamental functions. Keep practicing, and you will master this.

Common Traps & FAQs

Why can't I just integrate

(cos x - sin x)

from

0

\frac{3 π}{2}

directly?

That is a classic trap! If you integrate without the absolute value, the areas where the sine curve is above the cosine curve will be treated as negative, effectively cancelling out the positive areas. You would get a net displacement rather than the total geometric area. Always split the integral at the intersection points!

How do I know which function is on top in each interval?

Pick a test point! For the interval

[0, \frac{π}{4}]

, choose

x = 0

. At

x = 0

cos (0) = 1

and

sin (0) = 0

. Since

1 > 0

, the cosine curve is on top. Do this for every interval to ensure your subtraction order is correct.

Is there a symmetry I can use to save time?

Absolutely! Notice that the region from

0

\frac{π}{4}

and the region from

\frac{5 π}{4}

\frac{3 π}{2}

are symmetric. Calculating one and doubling it can save you precious minutes in the exam hall.

Animated Solution for Mathematics - Definite Integration: The area bounded by the curves $y = cos x$ and $y = sin x$ between the ordinates $x = 0$ and $x = \frac{3 π}{2}$ is

Select Answer:

Visualized Solution

The Curves and the Interval

Identifying the Upper and Lower Curves

Equating the Functions

Solving for Intersection Points

Splitting the Area

Setting up Integral for $A_{1}$

Calculating $A_{1}$

Setting up Integral for $A_{2}$

Calculating $A_{2}$

Setting up Integral for $A_{3}$

Calculating $A_{3}$

Total Area Calculation

The Sigma Insight: Area Bounded by Curves

The Geometric Dance of Sine and Cosine

Phase 1

Phase 2

Phase 3

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