
Assume that $f$ is continuous everywhere, then $\dfrac{1}{c}\int\limits_{ac}^{bc}{f\left( \dfrac{x}{c} \right)}dx=$
A. $\int\limits_{a}^{b}{f\left( \dfrac{x}{c} \right)}dx$
B. $\dfrac{1}{c}\int\limits_{a}^{b}{f\left( \dfrac{x}{c} \right)}dx$
C. $\int\limits_{a}^{b}{f\left( x \right)}dx$
D. None of these
Answer
164.1k+ views
Hint: In this question, we are to find the integral which is same as the given integral. For this, the variable substitution method is applied in the given integral. So, that the required integral will be obtained.
Formula Used:Definite integral:
Consider a function $f(x)$ is defined on $[a,b]$. If the integral of this function, $\int{f(x)dx=F(x)}$, then $F(b)-F(a)$ is called as the definite integral of the function $f(x)$ over $[a,b]$.
I.e.,$\int\limits_{a}^{b}{f(x)dx}=\left. F(x) \right|_{a}^{b}=F(b)-F(a)$
Here $a$ (lower limit) and $b$ (upper limit)
\[\int\limits_{a}^{b}{f(x)dx}=\int\limits_{a}^{b}{f(t)dt}\]
Some of the properties of the definite integrals are:
1) Interchanging the limits: \[\int\limits_{a}^{b}{f(x)dx=-}\int\limits_{b}^{a}{f(x)dx}\]
2) If $a3) $\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}$
4) $\int\limits_{-a}^{a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx}$ if $f(x)$ is an even function
$\int\limits_{-a}^{a}{f(x)dx}=0$ if $f(x)$ is an odd function
5) $\begin{align}
& \int\limits_{0}^{2a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx};\text{if }f(2a-x)=f(x) \\
& \text{ }=0\text{ if }f(2a-x)=-f(x) \\
\end{align}$
Complete step by step solution:Given integral is
$I=\dfrac{1}{c}\int\limits_{ac}^{bc}{f\left( \dfrac{x}{c} \right)}dx$
Consider $\dfrac{x}{c}=t$
So,
$\begin{align}
& x=ct \\
& \Rightarrow dx=cdt \\
\end{align}$
Then, the limits are
If $x=bc$ then $t=\dfrac{bc}{c}=b$
If $x=ac$ then $t=\dfrac{ac}{c}=a$
Then, the given integral become
\[\begin{align}
& I=\dfrac{1}{c}\int\limits_{ac}^{bc}{f\left( \dfrac{x}{c} \right)}dx \\
& \text{ }=\dfrac{1}{c}\int\limits_{a}^{b}{f\left( t \right)(c}dt) \\
& \text{ }=\dfrac{c}{c}\int\limits_{a}^{b}{f\left( t \right)}dt \\
& \text{ }=\int\limits_{a}^{b}{f\left( x \right)}dx \\
\end{align}\]
Option ‘C’ is correct
Note: Here, we applied substitution method to evaluate the integral. Here we may forget to change the limits. It is must that the limits should be altered as per the substituting variable. This method of solving an integral is an easy way of evaluating a definite integral.
Formula Used:Definite integral:
Consider a function $f(x)$ is defined on $[a,b]$. If the integral of this function, $\int{f(x)dx=F(x)}$, then $F(b)-F(a)$ is called as the definite integral of the function $f(x)$ over $[a,b]$.
I.e.,$\int\limits_{a}^{b}{f(x)dx}=\left. F(x) \right|_{a}^{b}=F(b)-F(a)$
Here $a$ (lower limit) and $b$ (upper limit)
\[\int\limits_{a}^{b}{f(x)dx}=\int\limits_{a}^{b}{f(t)dt}\]
Some of the properties of the definite integrals are:
1) Interchanging the limits: \[\int\limits_{a}^{b}{f(x)dx=-}\int\limits_{b}^{a}{f(x)dx}\]
2) If $a
4) $\int\limits_{-a}^{a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx}$ if $f(x)$ is an even function
$\int\limits_{-a}^{a}{f(x)dx}=0$ if $f(x)$ is an odd function
5) $\begin{align}
& \int\limits_{0}^{2a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx};\text{if }f(2a-x)=f(x) \\
& \text{ }=0\text{ if }f(2a-x)=-f(x) \\
\end{align}$
Complete step by step solution:Given integral is
$I=\dfrac{1}{c}\int\limits_{ac}^{bc}{f\left( \dfrac{x}{c} \right)}dx$
Consider $\dfrac{x}{c}=t$
So,
$\begin{align}
& x=ct \\
& \Rightarrow dx=cdt \\
\end{align}$
Then, the limits are
If $x=bc$ then $t=\dfrac{bc}{c}=b$
If $x=ac$ then $t=\dfrac{ac}{c}=a$
Then, the given integral become
\[\begin{align}
& I=\dfrac{1}{c}\int\limits_{ac}^{bc}{f\left( \dfrac{x}{c} \right)}dx \\
& \text{ }=\dfrac{1}{c}\int\limits_{a}^{b}{f\left( t \right)(c}dt) \\
& \text{ }=\dfrac{c}{c}\int\limits_{a}^{b}{f\left( t \right)}dt \\
& \text{ }=\int\limits_{a}^{b}{f\left( x \right)}dx \\
\end{align}\]
Option ‘C’ is correct
Note: Here, we applied substitution method to evaluate the integral. Here we may forget to change the limits. It is must that the limits should be altered as per the substituting variable. This method of solving an integral is an easy way of evaluating a definite integral.
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