Question

What is the integration of $\sqrt x $ limits $1$ to $4$?
A. $0$
B. $\dfrac{\pi }{4}$
C. $\dfrac{{{\pi ^2}}}{4}$
D. $\dfrac{{{\pi ^2}}}{2}$

Answer 1

Hint: In the given question, we are provided a definite integral to solve. The given problem revolves around the concepts and properties of definite integration. The given question requires us to integrate a function of x with respect to x. After integrating the function, we will put the lower and upper limits into the function and find the difference of the values.

Complete step by step answer:
The given question requires us to evaluate a definite integral $\int\limits_1^4 {\sqrt x } $ in variable x consisting of an algebraic expression.So, the function to be integrated is $\sqrt x $, the upper limit of integration is $4$ and the lower limit is $1$.So, to evaluate the definite integral, we will first do the indefinite integration of the function and then put the upper limit and lower limit into the obtained function and compute their difference. So, we have,
$\int_1^4 {\sqrt x \,dx} $

Now, we know the power rule of integration. So, we have,
$\int {{x^n}\,dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$
Applying the power rule in the problem, we get,
$ \Rightarrow \int_1^4 {\sqrt x \,dx} = \int_1^4 {{{\left( x \right)}^{\dfrac{1}{2}}}\,dx} $
$ \Rightarrow \int_1^4 {\sqrt x \,dx} = \left[ {\dfrac{{{{\left( x \right)}^{1 + \dfrac{1}{2}}}}}{{\left( {1 + \dfrac{1}{2}} \right)}}} \right]_1^4$
Simplifying the expression, we get,
$ \Rightarrow \int_1^4 {\sqrt x \,dx} = \left[ {\dfrac{{2{x^{\dfrac{3}{2}}}}}{3}} \right]_1^4$

Substituting the limits into the function, we get,
$ \Rightarrow \int_1^4 {\sqrt x \,dx} = \left[ {\dfrac{{2{{\left( 4 \right)}^{\dfrac{3}{2}}}}}{3} - \dfrac{{2{{\left( 1 \right)}^{\dfrac{3}{2}}}}}{3}} \right]$
Computing the powers and simplifying the value of integral, we get,
$ \Rightarrow \int_1^4 {\sqrt x \,dx} = \left[ {\dfrac{{2 \times 8}}{3} - \dfrac{{2 \times 1}}{3}} \right]$
$ \Rightarrow \int_1^4 {\sqrt x \,dx} = \left[ {\dfrac{{16}}{3} - \dfrac{2}{3}} \right]$
Simplifying the calculations, we get the value of definite integral as,
$ \therefore \int_1^4 {\sqrt x \,dx} = \dfrac{{14}}{3}$

Therefore, we get the value of the integral $\int\limits_1^4 {\sqrt x } $ as $\dfrac{{14}}{3}$.

Note:Indefinite integration gives us a family of curves. Definite integral gives us a numeric value after putting in the lower and upper limits into the function obtained after integration. An arbitrary constant has to be added in indefinite integration but the definite integration gives us a particular function and thus no constant is to be added.