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Question: How do you evaluate \( \int_0^1 {\sqrt {x - {x^2}} \,dx} \) ?...

How do you evaluate ∫01x−x2 dx\int_0^1 {\sqrt {x - {x^2}} \,dx} ?

Explanation

Solution

Hint : Here in this question, applying the integration directly to the function may be complicated. So to solve this question we use the formula ∫2au−u2 du=u−a22au−2+a22cos⁡−1(a−ua)+c\int {\sqrt {2au - {u^2}} \,du = \dfrac{{u - a}}{2}} \sqrt {2au{ - ^2}} + \dfrac{{{a^2}}}{2}{\cos ^{ - 1}}\left( {\dfrac{{a - u}}{a}} \right) + c where 2au−u2\sqrt {2au - {u^2}} , a>0a > 0 then after integrating and applying the limit value of integration and then we simplify the function to get the required result.

Complete step-by-step answer :
The given integral in the question is a definite integral defined as a definite integral characterized by upper and lower limits. Moreover, the reason why it is called definite is because it provides a definite answer at the end of the problem.
To evaluate the given integral using the formula
∫2au−u2 du=u−a22au−u2+a22cos⁡−1(a−ua)+c\int {\sqrt {2au - {u^2}} \,du = \dfrac{{u - a}}{2}} \sqrt {2au{ -u ^2}} + \dfrac{{{a^2}}}{2}{\cos ^{ - 1}}\left( {\dfrac{{a - u}}{a}} \right) + c where 2au−u2\sqrt {2au - {u^2}} , a>0a > 0
Where c is an integrating constant but in the definite integral we are not put any constant in this case we substitute the given limit points to the variable on simplification.
Compare the formula with given integral i.e., 2au−u22au - {u^2} with x−x2x - {x^2} and we see that a=12a = \dfrac{1}{2} provides the solution that we seek; thus, with a=12a = \dfrac{1}{2} we get using the formula provided, that:
Here limit point means variable x varies from 0 to 1 i.e., x:0→1x:0 \to 1
Consider ∫01x−x2 dx\int_0^1 {\sqrt {x - {x^2}} \,dx}
Now, integrating with respect to x by using the above formula, then
⇒   I=[x−122(12)x−x2+(12)22cos⁡−1(12−x12)]01\Rightarrow \,\,\,I = \left[ {\dfrac{{x - 1}}{2}\sqrt {2\left( {\dfrac{1}{2}} \right)x - {x^2}} + \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^2}}}{2}{{\cos }^{ - 1}}\left( {\dfrac{{\dfrac{1}{2} - x}}{{\dfrac{1}{2}}}} \right)} \right]_0^1
⇒I=[x−12x−x2+(14)2cos⁡−1(1−2x212)]01\Rightarrow I = \left[ {\dfrac{{x - 1}}{2}\sqrt {x - {x^2}} + \dfrac{{\left( {\dfrac{1}{4}} \right)}}{2}{{\cos }^{ - 1}}\left( {\dfrac{{\dfrac{{1 - 2x}}{2}}}{{\dfrac{1}{2}}}} \right)} \right]_0^1
⇒I=[x−12x−x2+18cos⁡−1(1−2x)]01\Rightarrow I = \left[ {\dfrac{{x - 1}}{2}\sqrt {x - {x^2}} + \dfrac{1}{8}{{\cos }^{ - 1}}\left( {1 - 2x} \right)} \right]_0^1
Now applying the upper and lower limit to the variable x, then
⇒I=(1−121−12+18cos⁡−1(1−2(1)))−(0−120−02+18cos⁡−1(1−2(0)))\Rightarrow I = \left( {\dfrac{{1 - 1}}{2}\sqrt {1 - {1^2}} + \dfrac{1}{8}{{\cos }^{ - 1}}\left( {1 - 2(1)} \right)} \right) - \left( {\dfrac{{0 - 1}}{2}\sqrt {0 - {0^2}} + \dfrac{1}{8}{{\cos }^{ - 1}}\left( {1 - 2\left( 0 \right)} \right)} \right) ⇒I=[0+18cos⁡−1(−1)−0−18cos⁡−1(1)]\Rightarrow I = \left[ {0 + \dfrac{1}{8}{{\cos }^{ - 1}}\left( { - 1} \right) - 0 - \dfrac{1}{8}{{\cos }^{ - 1}}\left( 1 \right)} \right]
⇒I=[0+18cos⁡−1(−1)−0−18cos⁡−1(1)]\Rightarrow I = \left[ {0 + \dfrac{1}{8}{{\cos }^{ - 1}}\left( { - 1} \right) - 0 - \dfrac{1}{8}{{\cos }^{ - 1}}\left( 1 \right)} \right]
The value of cos⁡−1(1)=0{\cos ^{ - 1}}\left( 1 \right) = 0 and cos⁡−1(−1)=π{\cos ^{ - 1}}\left( { - 1} \right) = \pi , then
⇒I=[18π−180]\Rightarrow I = \left[ {\dfrac{1}{8}\pi - \dfrac{1}{8}0} \right]
⇒I=π8\Rightarrow I = \dfrac{\pi }{8}
Hence, the value of the given integral function ∫01x−x2 dx\int_0^1 {\sqrt {x - {x^2}} \,dx} is π8\dfrac{\pi }{8} .
So, the correct answer is “ ∫01x−x2 dx\int_0^1 {\sqrt {x - {x^2}} \,dx} is π8\dfrac{\pi }{8} ”.

Note : In this type of question sometimes they will not mention integrating the function. If the question involves ∫\int {} this symbol means we have to integrate the function. In integration we have two kinds of integral namely, definite integral and indefinite integral. In the definite integral we have limit points. In an indefinite integral we don’t have limit points.