Question

If \[\int_a^b {{x^3}} dx = 0\] and \[\int_a^b {{x^2}} dx = \dfrac{2}{3}\]. What are the value of \[a\] and \[b\] ?
A.1,1
B.-1,-1
C.1,-1
D.-1,1

Answer 1

Hint: We will solve the first integration \[\int_a^b {{x^3}} dx = 0\] and derive a condition. Then we will solve the second integration\[\int_a^b {{x^2}} dx = \dfrac{2}{3}\]. In the solution of the second integration, we will put the value of \[b\] which we get from the solution of the first integration.

Formula Used: \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c\]
\[\int\limits_a^b {{x^n}dx} = \dfrac{b^{n + 1}}{n + 1} - \dfrac{a^{n + 1}}{n + 1}\]

Complete answer: We know the formula of power rule in the integration
We know the formula of power rule in the integration
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c\]
The formula of the definite integral is \[\int\limits_a^b {{x^n}dx} = \dfrac{{{b^{n + 1}}}}{{n + 1}} - \dfrac{{{a^{n + 1}}}}{{n + 1}}\]
The first integration is
\[\int_a^b {{x^3}} dx = 0\]
Now we will apply the formula\[\int\limits_a^b {{x^n}dx} = \dfrac{{{b^{n + 1}}}}{{n + 1}} - \dfrac{{{a^{n + 1}}}}{{n + 1}}\], here \[n = 3\].
  \[\dfrac{{{b^{3 + 1}}}}{{3 + 1}} - \dfrac{{{a^{3 + 1}}}}{{3 + 1}} = 0\]
 \[ \Rightarrow \dfrac{{{b^4}}}{4} - \dfrac{{{a^4}}}{4} = 0\]
\[ \Rightarrow {b^4} - {a^4} = 0\]
\[ \Rightarrow \left( {{b^2} - {a^2}} \right)\left( {{b^2} + {a^2}} \right) = 0\]
Either or,
\[\left( {{b^2} - {a^2}} \right) = 0\] or \[\left( {{b^2} + {a^2}} \right) = 0\]
\[\left( {b - a} \right)\left( {b + a} \right) = 0\]
Either or
\[\begin{array}{l}b - a = 0\\ \Rightarrow b = a\end{array}\] \[\begin{array}{l}b + a = 0\\ \Rightarrow b = - a\end{array}\]
If \[b = a\], then the value of \[\int_a^b {{x^2}} dx\] is also zero. But \[\int_a^b {{x^2}} dx \ne 0\]. Thus \[b = - a\].
Now we will solve the second integration
\[\int_a^b {{x^2}} dx = \dfrac{2}{3}\]
Apply the formula \[\int\limits_a^b {{x^n}dx} = \dfrac{{{b^{n + 1}}}}{{n + 1}} - \dfrac{{{a^{n + 1}}}}{{n + 1}}\], here \[n = 2\].
\[ \Rightarrow \dfrac{{{b^{2 + 1}}}}{{2 + 1}} - \dfrac{{{a^{2 + 1}}}}{{2 + 1}} = \dfrac{2}{3}\]
\[ \Rightarrow \dfrac{{{b^3}}}{3} - \dfrac{{{a^3}}}{3} = \dfrac{2}{3}\]
\[ \Rightarrow {b^3} - {a^3} = \dfrac{2}{3} \cdot 3\]
\[ \Rightarrow {b^3} - {a^3} = 2\]
\[ \Rightarrow {\left( { - a} \right)^3} - {a^3} = 2\]
\[ \Rightarrow - {a^3} - {a^3} = 2\]
\[ \Rightarrow - 2{a^3} = 2\]
\[ \Rightarrow {a^3} = - 1\]
\[ \Rightarrow a = - 1\]
Now we will put \[a = - 1\] in \[b = - a\].
\[\begin{array}{c}b = - \left( { - 1} \right)\\ = 1\end{array}\]
The value of \[a\] and \[b\] are -1 and 1 respectively.

Option ‘D’ is correct

Note: The cube root of a negative number is always a negative number and the cube root of a positive number is a positive number.