Problem 20 - Basic Engineering Mathematics

Question

Find the area enclosed by the curve \(y=\sin 2 x\), the \(x\)-axis and the ordinates \(x=0\) and \(x=\frac{\pi}{3}\),

Step-by-Step Solution

Answer

The area is \(\frac{3}{4}\).

Step 1: Understand the Problem

We need to find the area under the curve of the function \(y = \sin 2x\) from \(x=0\) to \(x=\frac{\pi}{3}\). This area is bounded above by the curve, below by the \(x\)-axis, and the vertical lines or the ordinates at \(x=0\) and \(x=\frac{\pi}{3}\).

Step 2: Set Up the Integral

The area under the curve \(y = \sin 2x\) from \(x=0\) to \(x=\frac{\pi}{3}\) is given by the definite integral:\[A = \int_{0}^{\frac{\pi}{3}} \sin 2x\, dx\].

Step 3: Find the Antiderivative

To solve the integral, we first find the antiderivative of \(\sin 2x\). The antiderivative of \(\sin(kx)\) is \(-\frac{1}{k}\cos(kx)\). Therefore, the antiderivative of \(\sin 2x\) is:\[-\frac{1}{2}\cos 2x\].

Step 4: Evaluate the Definite Integral

Substitute the limits of integration into the antiderivative. Evaluating \(-\frac{1}{2}\cos 2x\) from \(x=0\) to \(x=\frac{\pi}{3}\), we get:\[-\frac{1}{2}\cos(2\times\frac{\pi}{3}) + \frac{1}{2}\cos(2\times 0)\].

Step 5: Simplify the Expression

Calculate the cosine values:\(\cos(\frac{2\pi}{3}) = -\frac{1}{2}\) and \(\cos(0) = 1\). Substitute these values into the expression:\[-\frac{1}{2}(-\frac{1}{2}) + \frac{1}{2}(1) = \frac{1}{4} + \frac{1}{2}\].

Step 6: Add and Finalize the Area

Now add the results from the expression:\(\frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4}\). Therefore, the area enclosed by the curve is \(\frac{3}{4}\).

Key Concepts

Definite Integral

The definite integral is a fundamental concept in calculus, representing the area under a curve between two points along the x-axis. It is expressed in the form of \[ \int_{a}^{b} f(x) \, dx,\]where \(a\) and \(b\) are the limits of integration, representing the interval over which you calculate the area. In our exercise, we are looking at the function \(y = \sin 2x\) from \(x = 0\) to \(x = \frac{\pi}{3}\).

• Upper Limit: \(x=\frac{\pi}{3}\)
• Lower Limit: \(x=0\)
• Integral Function: \(f(x) = \sin 2x\)

By evaluating this integral, we find the exact area enclosed by the curve above the x-axis and between the given vertical lines. This process involves finding the antiderivative and then calculating the difference between its values at the bounds.

Trigonometric Functions

Trigonometric functions such as sine and cosine are essential in calculus, especially when dealing with periodic functions. In this exercise, the function \(y = \sin 2x\) is a trigonometric function that oscillates between -1 and 1. This function modifies the standard sine function by a frequency of 2.

• \(\sin(2x)\) oscillates twice as fast as \(\sin(x)\).
• The period of \(\sin(2x)\) is \(\frac{\pi}{2}\), as opposed to the usual interval of \(2\pi\) for \(\sin(x)\).
• The amplitude of \(\sin(2x)\) remains 1, which is the peak value of the function.

Understanding the behavior of \(\sin 2x\) helps us visualize the curve's path and its enclosed area. This is particularly important since trigonometric integrals often involve using identities or patterns from trigonometric graphs.

Antiderivative

Finding the antiderivative is a crucial step in solving definite integrals. The antiderivative of a function \(f(x)\) is another function \(F(x)\) such that its derivative is equal to \(f(x)\). For trigonometric functions like sine and cosine, this inverse operation involves reversing the process of differentiation.
In this case, we need to compute the antiderivative of \(\sin 2x\). The rule to remember here is:\[ \int \sin(kx) \, dx = -\frac{1}{k}\cos(kx).\]

• For \(\sin 2x\), \(k = 2\), so the antiderivative becomes \(-\frac{1}{2}\cos 2x\).

After finding the antiderivative, you substitute the limits of integration to determine the definite integral. This substitution and simplification gives you the area under the curve between the two specified x-values.