Courses
Courses for Kids
Free study material
Free LIVE classes
More
LIVE
Join Vedantu’s FREE Mastercalss

# If $\int\limits_{a}^{b}{{{x}^{3}}dx}=0$ and $\int\limits_{a}^{b}{{{x}^{2}}dx}=\dfrac{2}{3}$,then what are the values of a and b respectively?(a) $-1,1$(b) $1,1$(c) $0,0$(d) $2,-2$

Verified
362.7k+ views
Hint: We are given two basic integrals which, after solving, will give us two equations having
variables $'a'$ and $'b'$. Solving these two equations by substitution method, we can find out the
value of $'a'$ and $'b'$.

Before proceeding with the question, we will first discuss the formula which is required to
solve this question.
We have a formula in integration which can be used to integrate polynomial,

\begin{align} & \int\limits_{a}^{b}{{{x}^{n}}dx}=\left[ \dfrac{{{x}^{n+1}}}{n+1} \right]_{a}^{b} \\ & \Rightarrow \int\limits_{a}^{b}{{{x}^{n}}dx}=\dfrac{1}{n+1}\left( {{b}^{n+1}}-{{a}^{n+1}} \right)...........\left( 1 \right) \\ \end{align}

In the question, it is given $\int\limits_{a}^{b}{{{x}^{3}}dx}=0...........\left( 2 \right)$.
Substituting $n=3$ in equation $\left( 1 \right)$, we get,
\begin{align} & \int\limits_{a}^{b}{{{x}^{3}}dx}=\left[ \dfrac{{{x}^{3+1}}}{3+1} \right]_{a}^{b} \\ & \Rightarrow \int\limits_{a}^{b}{{{x}^{3}}dx}=\left[ \dfrac{{{x}^{4}}}{4} \right]_{a}^{b} \\ & \Rightarrow \int\limits_{a}^{b}{{{x}^{3}}dx}=\dfrac{1}{4}\left( {{b}^{4}}-{{a}^{4}} \right)..........\left( 3 \right) \\ \end{align}

Substituting $\int\limits_{a}^{b}{{{x}^{3}}dx}$ from equation $\left( 3 \right)$ in equation $\left( 2 \right)$, we get,
\begin{align} & \dfrac{1}{4}\left( {{b}^{4}}-{{a}^{4}} \right)=0 \\ & \Rightarrow {{b}^{4}}-{{a}^{4}}=0 \\ & \Rightarrow {{a}^{4}}={{b}^{4}} \\ \end{align}
$\Rightarrow a=+b$ or $a=-b.........\left( 4 \right)$
Also, it is given in the question $\int\limits_{a}^{b}{{{x}^{2}}dx}=\dfrac{2}{3}...........\left( 5 \right)$
Substituting $n=2$ in equation $\left( 1 \right)$, we get,
\begin{align} & \int\limits_{a}^{b}{{{x}^{2}}dx}=\left[ \dfrac{{{x}^{2+1}}}{2+1} \right]_{a}^{b} \\ & \Rightarrow \int\limits_{a}^{b}{{{x}^{2}}dx}=\left[ \dfrac{{{x}^{3}}}{3} \right]_{a}^{b} \\ & \Rightarrow \int\limits_{a}^{b}{{{x}^{2}}dx}=\dfrac{1}{3}\left( {{b}^{3}}-{{a}^{3}} \right)..........\left( 6 \right) \\ \end{align}
Substituting $\int\limits_{a}^{b}{{{x}^{2}}dx}$ from equation $\left( 6 \right)$ in equation $\left( 5 \right)$, we get,
$\dfrac{1}{3}\left( {{b}^{3}}-{{a}^{3}} \right)=\dfrac{2}{3}$
Cancelling $3$ on both the sides of the equality in the above equation, we get,
${{b}^{3}}-{{a}^{3}}=2..........\left( 7 \right)$
Substituting $a$ from equation $\left( 4 \right)$ in equation $\left( 7 \right)$, we get,
${{b}^{3}}-{{\left( +b \right)}^{3}}=2$ or ${{b}^{3}}-{{\left( -b \right)}^{3}}=2$
$\Rightarrow {{b}^{3}}-{{b}^{3}}=2$ or ${{b}^{3}}+{{b}^{3}}=2$
$0=2........\left( 8 \right)$ or $2{{b}^{3}}=2.........\left( 9 \right)$
We can conclude that equation $\left( 8 \right)$ is invalid because $0$ can never be equal to
$2$.Therefore, we will find our solution from equation $\left( 9 \right)$ only.
\begin{align} & 2{{b}^{3}}=2 \\ & \Rightarrow {{b}^{3}}=1 \\ & \Rightarrow b=1..........\left( 10 \right) \\ \end{align}

We can find $a$ by substituting the equation $\left( 10 \right)$ in equation $\left( 4 \right)$. Since we
had ignored equation $\left( 8 \right)$, we will not substitute equation $\left( 10 \right)$ in $a=+b$.
We will substitute the equation $\left( 10 \right)$ only in $a=-b$.
Substituting $b=1$ from equation $\left( 10 \right)$ in $a=-b$, we get,
$a=-1..........\left( 11 \right)$
Hence, from equation $\left( 10 \right)$ and equation $\left( 11 \right)$, we obtain $a=-1,b=1$.

Therefore the correct answer is option (a).

Note: Sometimes in such types of questions, we may apply the incorrect formula of the integration that we discussed in equation $\left( 1 \right)$. We sometimes apply it as
$\int\limits_{a}^{b}{{{x}^{n}}dx}=\dfrac{1}{n-1}\left( {{b}^{n-1}}-{{a}^{n-1}} \right)$ instead of
$\int\limits_{a}^{b}{{{x}^{n}}dx}=\dfrac{1}{n+1}\left( {{b}^{n+1}}-{{a}^{n+1}} \right)$ because we convert $n$ to $n-1$ in differentiation.
Last updated date: 25th Sep 2023
Total views: 362.7k
Views today: 10.62k