
$\int\limits_{0}^{1}{f(1-x)dx}$ has the same value as the integral
A. \[\int\limits_{0}^{1}{f(x)dx}\]
B. \[\int\limits_{0}^{1}{f(-x)dx}\]
C. \[\int\limits_{0}^{1}{f(x-1)dx}\]
D. \[\int\limits_{-1}^{1}{f(x)dx}\]
Answer
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Hint: In this question, we are to find the integral which is the 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 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
$I=\int\limits_{0}^{1}{f(1-x)dx}$
Substituting $1-x=t$
So, we can write $x=1-t\Rightarrow dx=-dt$
Then, the limits become
If $x=0$, then $t=1-0=1$
If $x=1$, then $t=1-1=0$
On substituting all these in the given integral, we get
$\begin{align}
& I=\int\limits_{0}^{1}{f(1-x)dx} \\
& \text{ }=\int\limits_{1}^{0}{f(t)(-dt)} \\
& \text{ }=-\int\limits_{0}^{1}{f(t)(-dt)} \\
& \text{ }=\int\limits_{0}^{1}{f(t)dt} \\
\end{align}$
Thus, we can write
$\int\limits_{0}^{1}{f(t)dt}=\int\limits_{0}^{1}{f(x)dx}$
Therefore, the integral that is the same as the given integral is
$\int\limits_{0}^{1}{f(1-x)dx}=\int\limits_{0}^{1}{f(x)dx}$
Option ‘A’ is correct
Note: Here, we applied the substitution method to evaluate the integral. Here we may forget to change the limits. It is just 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 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
$I=\int\limits_{0}^{1}{f(1-x)dx}$
Substituting $1-x=t$
So, we can write $x=1-t\Rightarrow dx=-dt$
Then, the limits become
If $x=0$, then $t=1-0=1$
If $x=1$, then $t=1-1=0$
On substituting all these in the given integral, we get
$\begin{align}
& I=\int\limits_{0}^{1}{f(1-x)dx} \\
& \text{ }=\int\limits_{1}^{0}{f(t)(-dt)} \\
& \text{ }=-\int\limits_{0}^{1}{f(t)(-dt)} \\
& \text{ }=\int\limits_{0}^{1}{f(t)dt} \\
\end{align}$
Thus, we can write
$\int\limits_{0}^{1}{f(t)dt}=\int\limits_{0}^{1}{f(x)dx}$
Therefore, the integral that is the same as the given integral is
$\int\limits_{0}^{1}{f(1-x)dx}=\int\limits_{0}^{1}{f(x)dx}$
Option ‘A’ is correct
Note: Here, we applied the substitution method to evaluate the integral. Here we may forget to change the limits. It is just 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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