Question

Find the value of $g(x + \pi )$ in terms of $g(x)$and $g(\pi )$, if $g(x) = \int\limits_0^x {\cos 4tdt}$.
(A) $\dfrac{{g(x)}}{{g(\pi )}}$
(B) $g(x) + g(\pi )$
(C) ) $g(x) - g(\pi )$
(D) $g(\pi )$

Answer 1

For this question the limits is given as $0 \to x + \pi$ because we find $g(x + \pi )$ and then we break the limit into $0 \to x$ and $x \to x + \pi$.Now we find the value of ${l_1} = \int\limits_x^{x + \pi } {\cos 4tdt}$ now by using definite integrals we get the value of $g(\pi )$. After this now we find the value in terms of $g(x)$ and $g(\pi )$.

Complete step-by-step answer:
According to the question we find the value of $g(x + \pi )$ with the help of $g(x) = \int\limits_0^x {\cos 4tdt}$
Now we find the value of $g(x + \pi )$ in terms of $g(x)$ and $g(\pi )$,
$\Rightarrow g(x + \pi ) = \int\limits_{t = 0}^{t = x + \pi } {\cos 4tdt}$
Now we take the limits as $0 \to x$ and $x \to x + \pi$, we get
$\Rightarrow g(x + \pi ) = \int\limits_0^x {\cos 4tdt} + \int\limits_x^{x + \pi } {\cos 4tdt}$
$\Rightarrow g(x + \pi ) = g(x) + {l_1}$
As $\int\limits_0^x {\cos 4tdt} = g(x)$ and let ${l_1} = \int\limits_x^{x + \pi } {\cos 4tdt}$
By using definite integral we get the value of ${l_1}$ as
${l_1} = \int\limits_0^\pi {\cos 4tdt} \\\ = g(\pi ) \\\$
Also we know that $\int \cos at =\dfrac{\sin at}{a}$
Using this formula and applying the limit values we get,

$${l_1} = \left( {\dfrac{{\sin 4t}}{4}} \right)_0^\pi \\\ = \left( {\dfrac{{\sin 4\pi }}{4} - \dfrac{{\sin 0}}{4}} \right) \\\ = 0 \\\$$

Hence the value of ${l_1}$ is 0.
Now as we know that $g(\pi ) = {l_1} = 0$,
So, the value of $g(x + \pi )$ does not depend upon $g(\pi )$
Hence we can add the value of $g(\pi )$ to $g(x)$
Hence $g(x + \pi ) = g(x) + g(\pi )$

So, the correct answer is “Option B”.

Additional Information: Integrals: In mathematics an object that can be interpreted or represented as an area or a generalization of area is termed as integrals. Integrals along with derivatives, are the fundamental objects of calculus.
Definite Integral: A Definite Integral is an integral which has start and end values. In other words we can say that definite integrals are bounded into limits.

Note: Alternate Method: by this alternate method we can also find the value of $g(x + \pi )$
According to the question it is given that $g(x) = \int\limits_0^x {\cos 4tdt}$
Now according to the integrals the value of $g(x + \pi ) = \int\limits_0^{x + \pi } {\cos 4tdt}$
By solving it we get,
$\Rightarrow g(x + \pi ) = \int\limits_0^\pi {\cos 4tdt} + \int\limits_\pi ^{x + \pi } {\cos 4tdt}$
$\Rightarrow g(x + \pi ) = g(\pi ) + \int\limits_0^\pi {\cos 4(u + \pi )du} \\\ where,t = u + \pi \\\$
By solving the equation we get,
$g(x + \pi ) = g(x) + g(\pi )$