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Question: What is the integration of \(\sqrt x \) limits \(1\) to \(4\)? A. \(0\) B. \(\dfrac{\pi }{4}\) ...

What is the integration of x\sqrt x limits 11 to 44?
A. 00
B. π4\dfrac{\pi }{4}
C. π24\dfrac{{{\pi ^2}}}{4}
D. π22\dfrac{{{\pi ^2}}}{2}

Explanation

Solution

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 ∫14x\int\limits_1^4 {\sqrt x } in variable x consisting of an algebraic expression.So, the function to be integrated is x\sqrt x , the upper limit of integration is 44 and the lower limit is 11.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,
∫14x dx\int_1^4 {\sqrt x \,dx}

Now, we know the power rule of integration. So, we have,
∫xn dx=xn+1n+1+c\int {{x^n}\,dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c
Applying the power rule in the problem, we get,
⇒∫14x dx=∫14(x)12 dx\Rightarrow \int_1^4 {\sqrt x \,dx} = \int_1^4 {{{\left( x \right)}^{\dfrac{1}{2}}}\,dx}
⇒∫14x dx=[(x)1+12(1+12)]14\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,
⇒∫14x dx=[2x323]14\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,
⇒∫14x dx=[2(4)323−2(1)323]\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,
⇒∫14x dx=[2×83−2×13]\Rightarrow \int_1^4 {\sqrt x \,dx} = \left[ {\dfrac{{2 \times 8}}{3} - \dfrac{{2 \times 1}}{3}} \right]
⇒∫14x dx=[163−23]\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,
∴∫14x dx=143\therefore \int_1^4 {\sqrt x \,dx} = \dfrac{{14}}{3}

Therefore, we get the value of the integral ∫14x\int\limits_1^4 {\sqrt x } as 143\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.