# definite integral How to determine the value of definite integral of f(x)=x(1+x)^8, if x=0 to x=1?

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You may also use the following alternative method, such that:

`int x(1 + x)^8 dx = int x*(1 + x)^4*(1 + x)^3*(1 + x)dx`

You need to come up with the substitution, such that:

`(1 + x)^4 = u => 4(1 + x)^3dx = du => (1 + x)^3dx = (du)/4`

`1 + x = root(4) u => x = root(4) u - 1`

Changing the limits of integration yields:

`x = 0 => u = 1`

`x = 1 => u = 2^4 = 16`

Changing the variable yields:

`int_1^16 (root(4) u - 1)*u*root(4) u*(du)/4 = (1/4)int_1^16(u*root(4)(u^2) - u*root(4) u) du`

Using the property of linearity, you need to split the integral in two easier integrals, such that:

`int_1^16 (root(4) u - 1)*u*root(4) u*(du)/4 = (1/4)int_1^16 (u*root(4)(u^2)) du - (1/4)int_1^16 (u*root(4) u) du`

Converting the radical into a rational power yields:

`int_1^16 (root(4) u - 1)*u*root(4) u*(du)/4 = (1/4)int_1^16 u*u^(1/2) du - (1/4)int_1^16 u*u^(1/4) du`

`int_1^16 (root(4) u - 1)*u*root(4) u*(du)/4 = (1/10)(u^(5/2))|_1^16 - (1/9)u^(9/4)|_1^16`

Using the fundamental theorem of calculus, yields:

`int_1^16 (root(4) u - 1)*u*root(4) u*(du)/4 = (1/10)(2^10 - 1) - (1/9)(2^9 - 1)`

Hence, evaluating the given definite integral yields` int x(1 + x)^8 dx = (1/10)(2^10 - 1) - (1/9)(2^9 - 1).`

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To evaluate the definite integral, we'll apply Leibniz-Newton formula;

Int f(x)dx = F(b) - F(a)

To solve the integral, we'll change the variable:

1 + x = t

x = t -  1

We'll differentiate both sides:

dx = dt

Int f(x)dx = Int (t-1)*t^8dt

Note that it's much more easier to raise the a variable to a power then to apply Newton binomial.

Int (t-1)*t^8dt = Int t^9 dt - Int t^8 dt

Int (t-1)*t^8dt = t^10/10 - t^9/9 + C

Int f(x)dx = F(1) - F(0)

F(1) =  (1 + 1)^10/10 - (1 + 1)^9/9

F(1) = (9*2^10 - 10*2^9)/9*10

F(0) = 1/10 - 1/9

F(0) = -1/9*10

F(1) - F(0) = [2^9(18 - 10) + 1]/9*10

Int f(x)dx = F(1) - F(0) = (2^12+ 1)/90