Question

\[f(x)= 4x+3, if~ 1 \leq x \leq 2 \\ ~~~~~~~~= 3x+5, if~2< x \leq 4\],
then \[\int\limits_{1}^{4}{f(x)dx=}\]
A. \[80\]
B. \[20\]
C. \[-20\]
D. \[37\]

Answer 1

Hint: In this question, we are to find the given integral. Here the function in the given integral is provided with two different intervals. According to the interval, the function is to be selected. This is achieved by subdividing the integrals into two parts with two different intervals in such a way that we get the given function. I.e., The interval [1,4] is subdivided into [1,2] and (2,4]. Thus, by applying these intervals for the given integral, we get the required value.



Formula Used:Definite integral:
Consider a function $f(x)$ is defined on [a, b]. If the integral of this function, $\int{f(x)dx=F(x)}$, then $F(b)-F(a)$ is called the definite integral of the function$f(x)$ over [a, b].
I.e., $\int\limits_{a}^{b}{f(x)dx}=\left. F(x) \right|_{a}^{b}=F(b)-F(a)$
Here $a$(lower limit) and $b$(upper limit).
\[\int\limits_{a}^{b}{f(x)dx}=\int\limits_{a}^{b}{f(t)dt}\]

Some of the properties of the definite integrals are:
1) Interchanging the limits: \[\int\limits_{a}^{b}{f(x)dx=-}\int\limits_{b}^{a}{f(x)dx}\]

2) If $a
3) $\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}$

4) $\int\limits_{-a}^{a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx}$ if $f(x)$ is an even function
$\int\limits_{-a}^{a}{f(x)dx}=0 if f(x)$ is an odd function

5)$\begin{align} \int_{0}^{2a}{f(x)dx}=2 \int_{0}^{a}{f(x)dx};\text{if }f(2a-x)=f(x) \\ \text{ }=0\text{ if }f(2a-x)=-f(x) \end{align}$

6) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\sin x)}{f(\sin x)+f(\cos x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cos x)}{f(\sin x)+f(\cos x)}dx=\dfrac{\pi }{4}}}$

7) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\tan x)}{f(\tan x)+f(\cot x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cot x)}{f(\tan x)+f(\cot x)}dx=\dfrac{\pi }{4}}}$



Complete step by step solution:Given integral is
\[I=\int\limits_{1}^{4}{f(x)dx}\]
And the given function is
\[f(x)= 4x+3, if~ 1 \leq x \leq 2 \\ ~~~~~~~~= 3x+5, if~2< x \leq 4\],
then \[\int\limits_{1}^{4}{f(x)dx=}\]
So, the given integral is
\[\begin{align} & I=\int\limits_{1}^{4}{f(x)dx}=\int\limits_{1}^{2}{f(x)dx+\int\limits_{2}^{4}{f(x)dx}} \ \\ \Rightarrow I=\int\limits_{1}^{2}{(4x+3)dx+\int\limits_{2}^{4}{(3x+5)dx}} \end{align}\]
We know that \[\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}\]
Then,
\[\begin{align} \int\limits_{1}^{4}{f(x)dx=}\left[ \dfrac{4{{x}^{2}}}{2}+3x \right]_{1}^{2}+\left[ \dfrac{3{{x}^{2}}}{2}+5x \right]_{2}^{4} \\ =\left[ 2{{(2)}^{2}}+3(2) \right]-\left[ 2{{(1)}^{2}}+3(1) \right]+\left[ \dfrac{3{{(4)}^{2}}}{2}+5(4) \right]-\left[ \dfrac{3{{(2)}^{2}}}{2}+5(2) \right] \\ =14-5+44-16 \\ =37 \end{align}\]



Option ‘D’ is correct



Note: Here we need to remember that, the given function is continuous at the given intervals. Thus, we can directly substitute the respective function at the respective interval. We need to remember this, as per the intervals the function has to be chosen. After that, the function should be integrated within those limits. On evaluating those, we get the value of the integral.