Question

If f(x)=\left\\{ \begin{aligned} & 2{{x}^{2}}+1,\text{ }x\le 1 \\\ & 4{{x}^{3}}-1,\text{ }x\ge 1 \\\ \end{aligned} \right\\}, then $\int\limits_{0}^{2}{f(x)dx}$ is
(a) $\dfrac{47}{3}$
(b) $\dfrac{50}{3}$
(c) $\dfrac{1}{3}$
(d) $\dfrac{47}{2}$

Answer 1

Hint:Break the limit of the integral from 0 to 1 and from 1 to 2 and then put the value of the function under their respective domain and integrate. Solve the two integrals separately.

Complete step-by-step answer:

We have been given that the function ‘$2{{x}^{2}}+1$’ is defined only for the values of $x$ less than or equal to 1 and the function ‘$4{{x}^{3}}-1$’ is defined for the values of $x$ greater than or equal to 1. Therefore, let us break the given limit of the integral. The given integral can be written as:
$\int\limits_{0}^{2}{f(x)dx}=\int\limits_{0}^{1}{f(x)dx}+\int\limits_{1}^{2}{f(x)dx}$
Substituting the value of $f(x)$ under their respective integral domains we get,
$\int\limits_{0}^{1}{f(x)dx}+\int\limits_{1}^{2}{f(x)dx}=\int\limits_{0}^{1}{(2{{x}^{2}}+1)dx}+\int\limits_{1}^{2}{(4{{x}^{3}}-1)dx}$
Let us assume that the value of the above integral is ‘$I$’. Therefore,
$I=\int\limits_{0}^{1}{(2{{x}^{2}}+1)dx}+\int\limits_{1}^{2}{(4{{x}^{3}}-1)dx}$
We know that, $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$. Therefore, applying this rule in the above integral, we have,
$I=\left[ \dfrac{2{{x}^{3}}}{3}+x \right]_{0}^{1}+\left[ \dfrac{4{{x}^{4}}}{4}-x \right]_{1}^{2}$
Substituting the value of suitable limits we get,
$\begin{aligned} & I=\left[ \left( \dfrac{2\times {{1}^{3}}}{3}+1 \right)-\left( \dfrac{2\times {{0}^{3}}}{3}+0 \right) \right]+\left[ \left( \dfrac{4\times {{2}^{4}}}{4}-2 \right)-\left( \dfrac{4\times {{1}^{4}}}{4}-1 \right) \right] \\\ & =\left[ \dfrac{2}{3}+1 \right]+\left[ \left( 16-2 \right)-\left( 1-1 \right) \right] \\\ & =\dfrac{5}{3}+14 \\\ & =\dfrac{47}{3} \\\ \end{aligned}$
Hence, option (a) is the correct answer.

Note: One may note that we have broken the given integral into two parts. One integral having limits ranging from 0 to 1 and the other having limits ranging from 0 to 2. This step was necessary because the two given functions are defined for different ranges of ‘$x$’, so, we cannot integrate them under one integral sign having limit ranging from 0 to 2.