Question

The area bounded by $y = \dfrac{{\sin x}}{x}$ , $x$ -axis and the ordinates $x = 0,x = \dfrac{\pi }{4}$ is
A.$ = \dfrac{\pi }{4}$
B.$ < \dfrac{\pi }{4}$
C.$ > \dfrac{\pi }{4}$
D.$ < \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\tan x}}{x}} dx$

Answer 1

Hint: Area under the curve will be determined using integration of a given function in the range $(0,\dfrac{\pi }{4})$ so draw the required graph of it before starting solving. And expand the function up to infinite terms.

Complete Step-by-Step solution:
The area under the curve between two points can be determined by using definite integral between the two points.
First step is to draw the diagram for proper understanding,

We know the area under the curve is found by integrating the curve between given ordinates.
Here, we integrate f(x) from $x = 0$ & $x = \dfrac{\pi }{4}$ .
Required area $ = \int\limits_0^{\dfrac{\pi }{4}} {f(x)dx} $
$I = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\sin x}}{x}} dx$
We know the expansion of $\sin x$ .
$\sin x = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - \dfrac{{{x^7}}}{{7!}} + ......$
So,
$I = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{(x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - \dfrac{{{x^7}}}{{7!}} + ......)}}{x}dx} $
Each term is divided by ‘x’, we get
$I = \int\limits_0^{\dfrac{\pi }{4}} {(1 - \dfrac{{{x^2}}}{{3!}}} + \dfrac{{{x^4}}}{{5!}} - \dfrac{{{x^6}}}{{7!}} + .....)dx$
We know, $\int {{x^n}dx = } \dfrac{{{x^{n + 1}}}}{{n + 1}}$
$I = [x - \dfrac{{{x^3}}}{{3 \times 3!}} + \dfrac{{{x^5}}}{{5 \times 5!}} - \dfrac{{{x^7}}}{{7 \times 7!}} + ....]_0^{\dfrac{\pi }{4}}$
$I = [\dfrac{\pi }{4} - \dfrac{{{{(\dfrac{\pi }{4})}^3}}}{{3 \times 3!}} + \dfrac{{{{(\dfrac{\pi }{4})}^5}}}{{5 \times 5!}} - ....]$
Now,
We can observe that it must be less than $\dfrac{\pi }{4}$ as some terms are subtracted from $\dfrac{\pi }{4}$.
So, correct answer is <$\dfrac{\pi }{4}$
option B) is correct.

Note: We know that we cannot find the value of ‘x’ at 0 as it would become undetermined i.e. $(\dfrac{0}{0})$form. To find the value of $\dfrac{{\sin x}}{x}$ at $x = 0$ is 1, we use the limit here to find its value at $x = 0$ .
.. . $\dfrac{{\sin x}}{x}$ is approaching zero at infinity.
And series expansion of $\sin x$ can be determined using the Maclaurin or Laurent method.