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Q3:
Find the area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$.
Solution :
Given Variables & Initial Setup
The area $A$ bounded by the curve $f(x) = \sin x$ between the limits $x = a$ and $x = b$ is defined by the definite integral of the absolute value of the function:
$A = \int_{0}^{2\pi} | \sin x | \, dx$
Step 1: Identifying Interval Sign Changes
The function $y = \sin x$ changes sign at the roots of the equation $\sin x = 0$ within the interval $[0, 2\pi]$. The roots are $x = 0, \pi, 2\pi$.
- For $x \in [0, \pi]$, $\sin x \ge 0$. [Per the unit circle definition in the first and second quadrants].
- For $x \in [\pi, 2\pi]$, $\sin x \le 0$. [Per the unit circle definition in the third and fourth quadrants].
Step 2: Splitting the Integral
By the property of additivity of definite integrals [Definite Integral Property: $\int_a^c f(x)dx = \int_a^b f(x)dx + \int_b^c f(x)dx$], we split the integral at $x = \pi$ to handle the absolute value:
$A = \int_{0}^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} | \sin x | \, dx$
$A = \int_{0}^{\pi} \sin x \, dx - \int_{\pi}^{2\pi} \sin x \, dx$
Step 3: Performing the Integration
The antiderivative of $\sin x$ is $-\cos x$. [Per the Fundamental Theorem of Calculus].
Part 1: $\int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$
Part 2: $-\int_{\pi}^{2\pi} \sin x \, dx = -[-\cos x]_{\pi}^{2\pi} = [\cos x]_{\pi}^{2\pi} = (\cos 2\pi - \cos \pi) = (1 - (-1)) = 2$
Step 4: Summation of Absolute Areas
$A = 2 + 2 = 4$
Visual Representation of the Bounded Area
Final Solution: The area bounded by the curve is 4 square units.
More Questions from Class 12 Mathematics Application of Integrals Miscellaneous Exercise on Chapter 8
- Q1(i): Find the area under the given curves and given lines: (i) $y = x^2$, $x = 1$, $x = 2$ and $x$-axis
- Q1(ii): Find the area under the given curves and given lines: (ii) $y = x^4$, $x = 1$, $x = 5$ and $x$-axis
- Q2: Sketch the graph of $y = |x + 3|$ and evaluate $\int_{-6}^{0} |x + 3| dx$.
- Q4: Area bounded by the curve $y = x^3$, the $x$-axis and the ordinates $x = – 2$ and $x = 1$ is
- Q5: The area bounded by the curve $y = x | x |$, $x$-axis and the ordinates $x = – 1$ and $x = 1$ is given by [Hint : $y = x^2$ if $x > 0$ and $y = – x^2$ if $x < 0$].
CBSE Solutions for Class 12 Mathematics Application of Integrals
Chapters in CBSE - Class 12 Mathematics
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