# Evaluate the value of the integral $\int_0^{10} {\left[ {\dfrac{{{x^{10}}}}{{{{\left( {10 - x} \right)}^{10}} + {x^{10}}}}} \right]} dx$.

Last updated date: 26th Mar 2023

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Answer

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Hint: Here, we will be proceeding by using the property of the definite integral which is $\int_a^b {\left[ {f(x)} \right]} dx = \int_a^b {\left[ {f(a + b - x)} \right]} dx$ where $f(x)$ is any function of x.

Complete step-by-step answer:

Let the given integral be ${\text{I}} = \int_0^{10} {\left[ {\dfrac{{{x^{10}}}}{{{{\left( {10 - x} \right)}^{10}} + {x^{10}}}}} \right]} dx{\text{ }} \to {\text{(1)}}$

According to the property of definite integral, we have

$\int_a^b {\left[ {f(x)} \right]} dx = \int_a^b {\left[ {f(a + b - x)} \right]} dx$

Using the above property, the integral given in equation (1) becomes

\[

{\text{I}} = \int_0^{10} {\left[ {\dfrac{{{{\left( {10 + 0 - x} \right)}^{10}}}}{{{{\left[ {10 - \left( {10 + 0 - x} \right)} \right]}^{10}} + {{\left( {10 + 0 - x} \right)}^{10}}}}} \right]} dx = \int_0^{10} {\left[ {\dfrac{{{{\left( {10 - x} \right)}^{10}}}}{{{{\left[ {10 - 10 + x} \right]}^{10}} + {{\left( {10 - x} \right)}^{10}}}}} \right]} dx \\

{\text{I}} = \int_0^{10} {\left[ {\dfrac{{{{\left( {10 - x} \right)}^{10}}}}{{{x^{10}} + {{\left( {10 - x} \right)}^{10}}}}} \right]} dx{\text{ }} \to {\text{(2)}} \\

\]

By adding equations (1) and (2), we get

$

{\text{I}} + {\text{I}} = \int_0^{10} {\left[ {\dfrac{{{x^{10}}}}{{{{\left( {10 - x} \right)}^{10}} + {x^{10}}}}} \right]} dx + \int_0^{10} {\left[ {\dfrac{{{{\left( {10 - x} \right)}^{10}}}}{{{x^{10}} + {{\left( {10 - x} \right)}^{10}}}}} \right]} dx \\

\Rightarrow 2{\text{I}} = \int_0^{10} {\left[ {\dfrac{{{x^{10}}}}{{{{\left( {10 - x} \right)}^{10}} + {x^{10}}}} + \dfrac{{{{\left( {10 - x} \right)}^{10}}}}{{{x^{10}} + {{\left( {10 - x} \right)}^{10}}}}} \right]} dx = \int_0^{10} {\left[ {\dfrac{{{x^{10}} + {{\left( {10 - x} \right)}^{10}}}}{{{{\left( {10 - x} \right)}^{10}} + {x^{10}}}}} \right]} dx \\

\Rightarrow 2{\text{I}} = \int_0^{10} {\left( 1 \right)} dx = \left[ x \right]_0^{10} = \left[ {10 - 0} \right] = 10 \\

\Rightarrow {\text{I}} = 5 \\

$

So, the value of the integral $\int_0^{10} {\left[ {\dfrac{{{x^{10}}}}{{{{\left( {10 - x} \right)}^{10}} + {x^{10}}}}} \right]} dx$ is 5.

Note: In these type of problems, we somehow convert the complex function given in terms of x which is inside the integral (here it is $\dfrac{{{x^{10}}}}{{{{\left( {10 - x} \right)}^{10}} + {x^{10}}}}$) into a simpler function (here it comes out to be 1) using some property of the definite integral so that the integral of the function can be easily evaluated.

Complete step-by-step answer:

Let the given integral be ${\text{I}} = \int_0^{10} {\left[ {\dfrac{{{x^{10}}}}{{{{\left( {10 - x} \right)}^{10}} + {x^{10}}}}} \right]} dx{\text{ }} \to {\text{(1)}}$

According to the property of definite integral, we have

$\int_a^b {\left[ {f(x)} \right]} dx = \int_a^b {\left[ {f(a + b - x)} \right]} dx$

Using the above property, the integral given in equation (1) becomes

\[

{\text{I}} = \int_0^{10} {\left[ {\dfrac{{{{\left( {10 + 0 - x} \right)}^{10}}}}{{{{\left[ {10 - \left( {10 + 0 - x} \right)} \right]}^{10}} + {{\left( {10 + 0 - x} \right)}^{10}}}}} \right]} dx = \int_0^{10} {\left[ {\dfrac{{{{\left( {10 - x} \right)}^{10}}}}{{{{\left[ {10 - 10 + x} \right]}^{10}} + {{\left( {10 - x} \right)}^{10}}}}} \right]} dx \\

{\text{I}} = \int_0^{10} {\left[ {\dfrac{{{{\left( {10 - x} \right)}^{10}}}}{{{x^{10}} + {{\left( {10 - x} \right)}^{10}}}}} \right]} dx{\text{ }} \to {\text{(2)}} \\

\]

By adding equations (1) and (2), we get

$

{\text{I}} + {\text{I}} = \int_0^{10} {\left[ {\dfrac{{{x^{10}}}}{{{{\left( {10 - x} \right)}^{10}} + {x^{10}}}}} \right]} dx + \int_0^{10} {\left[ {\dfrac{{{{\left( {10 - x} \right)}^{10}}}}{{{x^{10}} + {{\left( {10 - x} \right)}^{10}}}}} \right]} dx \\

\Rightarrow 2{\text{I}} = \int_0^{10} {\left[ {\dfrac{{{x^{10}}}}{{{{\left( {10 - x} \right)}^{10}} + {x^{10}}}} + \dfrac{{{{\left( {10 - x} \right)}^{10}}}}{{{x^{10}} + {{\left( {10 - x} \right)}^{10}}}}} \right]} dx = \int_0^{10} {\left[ {\dfrac{{{x^{10}} + {{\left( {10 - x} \right)}^{10}}}}{{{{\left( {10 - x} \right)}^{10}} + {x^{10}}}}} \right]} dx \\

\Rightarrow 2{\text{I}} = \int_0^{10} {\left( 1 \right)} dx = \left[ x \right]_0^{10} = \left[ {10 - 0} \right] = 10 \\

\Rightarrow {\text{I}} = 5 \\

$

So, the value of the integral $\int_0^{10} {\left[ {\dfrac{{{x^{10}}}}{{{{\left( {10 - x} \right)}^{10}} + {x^{10}}}}} \right]} dx$ is 5.

Note: In these type of problems, we somehow convert the complex function given in terms of x which is inside the integral (here it is $\dfrac{{{x^{10}}}}{{{{\left( {10 - x} \right)}^{10}} + {x^{10}}}}$) into a simpler function (here it comes out to be 1) using some property of the definite integral so that the integral of the function can be easily evaluated.

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