Question

Find the value of \[\int\limits_{ - 1}^1 {\left| {1 - x} \right|dx} \].
A. \[ - 2\]
B. \[0\]
C. \[2\]
D. \[4\]

Answer 1

Hint: In this question, we need to solve the integration and find the value. For this, we need to use the concept of modulus and find the appropriate value of \[\left| {1 - x} \right|\] for solving the integration. Also, the integration can be solved by using upper and lower limit values.

Formula used: The integration of a function \[f\left( x \right)\] over the interval \[\left( {a,b} \right)\] is given by
\[\int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right)\]

Complete step-by-step solution: Let \[I = \int\limits_{ - 1}^1 {\left| {1 - x} \right|dx} \]
It is in the form of \[I = \int\limits_{ - 1}^1 {f\left( x \right)dx} \]
Here, \[f\left( x \right) = \left| {1 - x} \right|\]
The concept of modulus is given by
\[
  \left| x \right| = x;{\text{ x}} \geqslant {\text{0}} \\
  {\text{ = }} - {\text{x; x < 0}} \\
 \]
According to the concept of modulus, we get
So, \[\left| {1 - x} \right| = \left( {1 - x} \right)\]
\[I = \int\limits_{ - 1}^1 {\left( {1 - x} \right)dx} \]
We know that \[\int {kdx = kx + C} \] and \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} \] where, \[k\] is constant_.
By integrating, we get
\[I = \left[ {x - \dfrac{{{x^2}}}{2}} \right]_{ - 1}^1\]
By putting upper and lower limits for the expression in terms of \[x\], we get
\[I = \left[ {1 - \dfrac{1}{2} - \left( { - 1 - \dfrac{1}{2}} \right)} \right]\]
By simplifying further, we get
\[I = \left[ {1 - \dfrac{1}{2} + 1 + \dfrac{1}{2}} \right]\]
Hence, we get
\[I = 2\]
That means, \[\int\limits_{ - 1}^1 {\left| {1 - x} \right|dx = 2} \]
Therefore, the value of \[\int\limits_{ - 1}^1 {\left| {1 - x} \right|dx} \] is \[2\].

Therefore, the option (C) is correct.

Additional Information: The definite integral is defined as the difference in the integral values of a particular function for an upper and lower limit of the independent variable. The ultimate value of a definite integral can be calculated by subtracting the integral to the lower limit from the integral to the upper limit. When the lower and upper boundaries are constants, a definite integral indicates a number.

Note: Many students make mistakes in finding modulus and solving the integration part using upper and lower limits. A modulus function is one that returns the absolute value of a number or variable. Also, it is called an absolute value function. This function usually produces a positive result, regardless of the input. Mathematically, it is represented as \[y = \left| x \right|\]