
Find the value of \[\int_{ - 1}^1 {x\left| x \right|dx} \].
A. 1
B. 0
C. 2
D. – 2
Answer
162.9k+ views
Hint: In this question, we are given an integral as \[\int_{ - 1}^1 {x\left| x \right|dx} \]. First, we will eliminate the absolutes to get \[\int_{ - 1}^0 {x\left( { - x} \right)dx} + \int_0^1 {x \cdot xdx} \].
Now, we will integrate from -1 to 0 and then from 0 to 1. At last, we will put the limits value to get the final result.
Formula Used:1) First, we will use the formula \[\int_a^b {f\left( x \right)dx = } \int_a^c {f\left( x \right)dx} + \int_c^a {f\left( x \right)dx} ,\left( {a < c < b} \right)\] to solve the given question.
2) Put \[ - x\] into the function in place of x to determine if it is even or odd. The function is even if \[f(x)\] is obtained in its original form. If the original function is obtained in the negative form, the function is odd.
In other words, we can say that
If \[f( - x) = f(x)\], the function is even
If \[f( - x) = - f(x)\], the function is odd
Complete step by step solution:We have been given \[\int_{ - 1}^1 {x\left| x \right|dx} \].
Let us consider that \[I = \int_{ - 1}^1 {x\left| x \right|dx} \]
We will first eliminate the absolute value by breaking the integral into different integrals.
\[I = \int_{ - 1}^0 {x \cdot \left( { - x} \right)dx} + \int_0^1 {x \cdot xdx} \]
\[ \Rightarrow I = \int_{ - 1}^0 { - {x^2}dx} + \int_0^1 {{x^2}dx} \]
Now we will integrate with respect to \[x\] to get
\[ \Rightarrow I = \left. { - \dfrac{{{x^3}}}{3}} \right|_{ - 1}^0 + \left. {\dfrac{{{x^3}}}{3}} \right|_0^1\]
We will further put the limits in the above integral.
\[ \Rightarrow I = - \left[ {\dfrac{{{0^3}}}{3} - \dfrac{{{{( - 1)}^3}}}{3}} \right] + \left[ {\dfrac{{{1^3}}}{3} - \dfrac{{{0^3}}}{3}} \right]\]
\[ \Rightarrow I = - \dfrac{1}{3} + \dfrac{1}{3} = 0\]
Option ‘B’ is correct
Note: As the functions change, it is not essential to integrate taking limits directly \[-1\] to\[1\]. We must avoid errors and confusion while dealing with sign changes that occurred during integration. We need to adopt the identities that make the problems easier.
Now, we will integrate from -1 to 0 and then from 0 to 1. At last, we will put the limits value to get the final result.
Formula Used:1) First, we will use the formula \[\int_a^b {f\left( x \right)dx = } \int_a^c {f\left( x \right)dx} + \int_c^a {f\left( x \right)dx} ,\left( {a < c < b} \right)\] to solve the given question.
2) Put \[ - x\] into the function in place of x to determine if it is even or odd. The function is even if \[f(x)\] is obtained in its original form. If the original function is obtained in the negative form, the function is odd.
In other words, we can say that
If \[f( - x) = f(x)\], the function is even
If \[f( - x) = - f(x)\], the function is odd
Complete step by step solution:We have been given \[\int_{ - 1}^1 {x\left| x \right|dx} \].
Let us consider that \[I = \int_{ - 1}^1 {x\left| x \right|dx} \]
We will first eliminate the absolute value by breaking the integral into different integrals.
\[I = \int_{ - 1}^0 {x \cdot \left( { - x} \right)dx} + \int_0^1 {x \cdot xdx} \]
\[ \Rightarrow I = \int_{ - 1}^0 { - {x^2}dx} + \int_0^1 {{x^2}dx} \]
Now we will integrate with respect to \[x\] to get
\[ \Rightarrow I = \left. { - \dfrac{{{x^3}}}{3}} \right|_{ - 1}^0 + \left. {\dfrac{{{x^3}}}{3}} \right|_0^1\]
We will further put the limits in the above integral.
\[ \Rightarrow I = - \left[ {\dfrac{{{0^3}}}{3} - \dfrac{{{{( - 1)}^3}}}{3}} \right] + \left[ {\dfrac{{{1^3}}}{3} - \dfrac{{{0^3}}}{3}} \right]\]
\[ \Rightarrow I = - \dfrac{1}{3} + \dfrac{1}{3} = 0\]
Option ‘B’ is correct
Note: As the functions change, it is not essential to integrate taking limits directly \[-1\] to\[1\]. We must avoid errors and confusion while dealing with sign changes that occurred during integration. We need to adopt the identities that make the problems easier.
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