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Question: Evaluate the value of the integral \(\int_0^{10} {\left[ {\dfrac{{{x^{10}}}}{{{{\left( {10 - x} \rig...

Evaluate the value of the integral ∫010[x10(10−x)10+x10]dx\int_0^{10} {\left[ {\dfrac{{{x^{10}}}}{{{{\left( {10 - x} \right)}^{10}} + {x^{10}}}}} \right]} dx.

Explanation

Solution

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

Complete step-by-step answer:
Let the given integral be I=∫010[x10(10−x)10+x10]dx →(1){\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
∫ab[f(x)]dx=∫ab[f(a+b−x)]dx\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

I=∫010[(10+0−x)10[10−(10+0−x)]10+(10+0−x)10]dx=∫010[(10−x)10[10−10+x]10+(10−x)10]dx I=∫010[(10−x)10x10+(10−x)10]dx →(2)  {\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
I+I=∫010[x10(10−x)10+x10]dx+∫010[(10−x)10x10+(10−x)10]dx ⇒2I=∫010[x10(10−x)10+x10+(10−x)10x10+(10−x)10]dx=∫010[x10+(10−x)10(10−x)10+x10]dx ⇒2I=∫010(1)dx=[x]010=[10−0]=10 ⇒I=5  {\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 ∫010[x10(10−x)10+x10]dx\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 x10(10−x)10+x10\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.