What is the area of the region bounded by the above two curves and the lines x = 0 and ${\rm{x}} = \frac{{\rm{\pi }}}{4}$ ?

Question

Prepp Team · Accepted Answer

Finding Area Bounded by Sine and Cosine Curves

The question asks for the area of the region enclosed by the curves $y = \sin x$, $y = \cos x$, and the vertical lines $x = 0$ and $x = \frac{\pi}{4}$.

To find the area between two curves $f(x)$ and $g(x)$ from $x=a$ to $x=b$, we use the definite integral: \[ A = \int_{a}^{b} |f(x) - g(x)| dx \] First, we need to determine which function is greater over the given interval $[0, \frac{\pi}{4}]$. Let's consider the values of $\sin x$ and $\cos x$ in this interval:

At $x = 0$: $\sin(0) = 0$, $\cos(0) = 1$. Here, $\cos x > \sin x$.
At $x = \frac{\pi}{4}$: $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. Here, $\cos x = \sin x$.
For values of $x$ between $0$ and $\frac{\pi}{4}$, the graph of $y = \cos x$ is above the graph of $y = \sin x$. For instance, at $x = \frac{\pi}{6}$, $\sin(\frac{\pi}{6}) = \frac{1}{2}$ and $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. Since $\frac{\sqrt{3}}{2} > \frac{1}{2}$, $\cos x > \sin x$.

Thus, over the interval $[0, \frac{\pi}{4}]$, $\cos x \ge \sin x$. The area is given by the integral of the difference $\cos x - \sin x$ from $0$ to $\frac{\pi}{4}$.

The integral for the area is:

\[ A = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) dx \]

Now, we evaluate the definite integral:

The antiderivative of $\cos x$ is $\sin x$.

The antiderivative of $\sin x$ is $-\cos x$.

So, the integral is:

\[ \int (\cos x - \sin x) dx = \sin x - (-\cos x) + C = \sin x + \cos x + C \]

Now, we apply the limits of integration from $0$ to $\frac{\pi}{4}$:

\[ A = [\sin x + \cos x]_{0}^{\frac{\pi}{4}} \] \[ A = (\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4})) - (\sin(0) + \cos(0)) \]

Substitute the known values:

\[ \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \] \[ \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \] \[ \sin(0) = 0 \] \[ \cos(0) = 1 \]

Plugging these values into the expression for A:

\[ A = \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - (0 + 1) \] \[ A = \left(\frac{2}{\sqrt{2}}\right) - 1 \]

Simplify the fraction $\frac{2}{\sqrt{2}}$:

\[ \frac{2}{\sqrt{2}} = \frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2} \]

So, the area is:

\[ A = \sqrt{2} - 1 \]

This value corresponds to one of the given options.

Step-by-Step Area Calculation

Identify the curves $y = \sin x$ and $y = \cos x$ and the boundary lines $x=0$ and $x=\frac{\pi}{4}$.
Determine the interval of integration: $[0, \frac{\pi}{4}]$.
Determine which curve is above the other in the interval $[0, \frac{\pi}{4}]$. By checking points or knowing the graphs, we find $\cos x \ge \sin x$ for $x \in [0, \frac{\pi}{4}]$.
Set up the definite integral for the area: $A = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) dx$.
Find the antiderivative of the integrand $(\cos x - \sin x)$, which is $\sin x + \cos x$.
Evaluate the antiderivative at the upper and lower limits and subtract: $A = [\sin x + \cos x]_{0}^{\frac{\pi}{4}}$.
Calculate the value: $A = (\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4})) - (\sin(0) + \cos(0)) = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1)$.
Simplify the result: $A = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1$.

Summary of Functions and Bounds

Element	Description
Curve 1	$y = \sin x$
Curve 2	$y = \cos x$
Lower x-bound	$x = 0$
Upper x-bound	$x = \frac{\pi}{4}$
Dominant curve in $[0, \frac{\pi}{4}]$	$y = \cos x$

Understanding the Area Concept

The area between two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ is the integral of the absolute difference between the functions over that interval. This accounts for cases where the curves might cross within the interval, requiring the integral to be split into sub-intervals where the dominant function is consistent.

In this specific problem, the curves $y = \sin x$ and $y = \cos x$ intersect at $x = \frac{\pi}{4}$ within the interval $[0, \frac{\pi}{2}]$. However, our interval of interest is $[0, \frac{\pi}{4}]$, which ends exactly at their intersection point. Over $[0, \frac{\pi}{4}]$, $\cos x$ is always greater than or equal to $\sin x$, simplifying the calculation as we don't need the absolute value.

Revision Table: Key Concepts

Concept	Description
Area between curves	Integral of the difference between the upper and lower function over the interval.
Definite integral	Used to calculate the area under a curve or between curves over a specific interval.
Trigonometric functions	Functions like sine ($\sin x$) and cosine ($\cos x$) with periodic properties.
Limits of Integration	The x-values (a and b) defining the interval over which the integral is evaluated.

Additional Information: Properties of Sine and Cosine

The functions $y = \sin x$ and $y = \cos x$ are fundamental trigonometric functions. They have a period of $2\pi$.

The graph of $y = \cos x$ is a horizontal shift of the graph of $y = \sin x$, specifically $\cos x = \sin(x + \frac{\pi}{2})$.
They intersect when $\sin x = \cos x$, which occurs at $x = \frac{\pi}{4} + n\pi$ for integer $n$. The first intersection in the positive x-axis is at $x = \frac{\pi}{4}$.
In the interval $[0, \frac{\pi}{2}]$, $\cos x$ starts at its maximum (1 at $x=0$) and decreases to its minimum (0 at $x=\frac{\pi}{2}$), while $\sin x$ starts at its minimum (0 at $x=0$) and increases to its maximum (1 at $x=\frac{\pi}{2}$).
The value $\frac{\pi}{4}$ is where they intersect, and it's also the midpoint of the interval $[0, \frac{\pi}{2}]$.

Prepp Team · Answer

$\sqrt 2 + 1$

Prepp Team · Answer

$\sqrt 2$

Prepp Team · Answer

2