Question

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

Answer 1

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 :
$\int 1 dx = x + c$
$\int a dx = ax + c$
$\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$ ; $n \ne 1$
$\int {\sin x} dx = - \cos x + c$
$\int {\cos x} dx = \sin x + c$
$\int {{{\sec }^2}x} dx = \tan x + c$