Question

The area bounded by the curve $y = \cos ax$ in one arc of the curve

1. $2a$
2. $\dfrac{2}{a}$
3. $\dfrac{1}{a}$
4. $2{a^2}$

Answer 1

First of all draw the diagram of the given curve. Find the limits of $x$ for the required region. Next, use definite integration to find the area bounded by the curve $y = \cos ax$

Complete step-by-step answer:
We will start solving by drawing the given curve and finding the required region bounded by $y = \cos ax$ in one arc.
We have to calculate the area of the curve in an arc.
The given curve is $y = \cos ax$, one arc of the curve will be from $x = 0$ and $x = \dfrac{\pi }{{2a}}$.
We will integrate the curve $y = \cos ax$, where $0 \leqslant x \leqslant \dfrac{\pi }{{2a}}$
Now, we will definite integration to find the area:
The area bounded by the curve $y = f\left( x \right)$ , and the region is in between $x = a$ and $x = b$ is given by $A = \int_a^b {f\left( x \right)dx}$
For this question, $f\left( x \right) = \cos ax$ and the region is from $x = 0$ and $x = \dfrac{\pi }{{2a}}$.
Hence, we have, $A = \int_0^{\dfrac{\pi }{{2a}}} {\cos axdx}$
$A = \left[ {\dfrac{{\sin ax}}{a}} \right]_0^{\dfrac{\pi }{{2a}}}$
On substituting the limits, we get,
$A = \left[ {\dfrac{{\sin a\left( {\dfrac{\pi }{{2a}}} \right)}}{a}} \right] - \left[ {\dfrac{{\sin a\left( 0 \right)}}{a}} \right] \\\ A = \dfrac{1}{a}\left( {\left( {\sin \dfrac{\pi }{2}} \right) - \sin 0} \right) \\\ A = \dfrac{1}{a} \\\$
Hence, the area bounded by the curve $y = \cos ax$ in one arc of the curve $\dfrac{1}{a}$ square units.
Therefore, option C is correct.

Note: One has to learn all the formulas of integration for this type of question. Also, sketching the graph for the required region plays an important role. Be careful while substituting the limits as we have to put the upper limit first and then followed by the lower limit.