Solve : \[\int\limits_0^1 {x{{(1 - x)}^5}} dx\]

Hint: We have to integrate the given function \[x{(1 - x)^5}\] with respect to ‘ \[x\] ’ for the limits \[0\] to \[1\] . We solve this using integration by substitution method and using the various formulas of integration . First we change the terms of the integration by substituting the value of \[\left( {1 - x} \right)\] with a variable and by differentiating the terms we will change the value of the limits . And on further integration of the terms and then putting the values of the limits in the integrated terms we get the required solution for the given integral expression .

Complete step-by-step answer:
Given : \[\int\limits_0^1 {x{{(1 - x)}^5}} dx\]
Let \[I = \int\limits_0^1 {x{{(1 - x)}^5}} dx\]


We have to integrate \[I\] with respect to \[x\]
Put \[\left( {1 - x} \right) = t\]
\[x = (1 - t)\]
Differentiate \[t\] with respect to \[x\] , we get
(Derivative of \[constant = 0\])
(Derivative \[{x^n} = n{x^{n - 1}}\])

\[\left[ {0 - 1} \right]dx = dt\]
\[ - dx = dt\]
Putting values of the limits in \[x\] we get the new limits as :
When \[x = 1\] then \[t = 0\]
When \[x = 0\] then \[t = 1\]

Now , the integral becomes

\[I = \int\limits_1^0 { - (1 - t){t^5}} dt\]

On further simplifying , we get
\[I = \int\limits_1^0 { - ({t^5} - {t^6})} dt\]
We also know that the formula of integration for a variable is given as :
\[\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}\]

Using the formula of integration , we get
\[I = \mathop {\left[ {\dfrac{{ - {t^6}}}{6} + \dfrac{{{t^7}}}{7}} \right]}\nolimits_1^0 \]

Putting the values of the limits in the integral , we get
\[\Rightarrow I = \left[ { - 0 + 0} \right] - \left[ {\dfrac{{ - 1}}{6} + \dfrac{1}{7}} \right]\]
\[\Rightarrow I = \left[ {\dfrac{1}{6} - \dfrac{1}{7}} \right]\]

Taking L.C.M. and solving further , we get
\[\Rightarrow I = \left[ {\dfrac{{7 - 6}}{{42}}} \right]\]
\[\Rightarrow I = \left[ {\dfrac{1}{{42}}} \right]\]

Thus , the value of \[\int\limits_0^1 {x{{(1 - x)}^5}} dx\] is \[\dfrac{1}{{42}}\] .

Note: As the question was of definite integral that’s why we didn’t add an integral constant ‘\[a\]’ to the integration . Also , we got a fixed value for the integration . If the question would have been of indefinite integral then we would have added the integral constant to the final answer .
The formula of integration for various trigonometric terms are given as :
\[\int 1 dx = x + c\]
\[\int a dx = ax + c\]
\[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}\] ; \[n \ne 1\]
\[\int {\sin x} dx = - \cos x + c\]
\[\int {\cos x} dx = \sin x + c\]
\[\int {{{\sec }^2}x} dx = \tan x + c\]