
$\int\limits_{0}^{1}{f(1-x)dx}$ 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
232.8k+ views
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.
Recently Updated Pages
Geometry of Complex Numbers Explained

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Algebra Made Easy: Step-by-Step Guide for Students

Trending doubts
JEE Main 2026: Admit Card Out, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

Inductive Effect and Its Role in Acidic Strength

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

