Fun with Integration

Calculus Level 4

Solve your way through the following steps and submit the result of (c) as your answer.

(a) First, prove that for all values of n R n\in\mathbb{R}

0 1 ( n 2 x n + 2 n x n + x n 1 ) d x n . \int _{ 0 }^{ 1 }{ (n^{ 2 }x^{ n }+2nx^{ n }+x^{ n }-1) } dx\equiv n.

The logical reasoning behind this proof will help in parts (b) and (c).

(b) Sketch, on the same axes, the curves representing f ( x ) = 16 x 3 1 f(x)=16x^3-1 and x = 1. x=1.

(c) Express f ( x ) f(x) in the form x n ( n + 1 ) 2 1 , x^n (n+1)^2-1, and hence, without integrating or making use of any approximations, state the area enclosed between the two curves and the coordinate axes.

Note: This problem can easily be worked out with a calculator, online maths website or by integrating directly. But that will take all the fun out of it! The question is based on the typical calculus A-level syllabus. Please follow the rules and arrive to the answer by following all the below steps chronologically.


The answer is 3.

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1 solution

Vincent Moroney
Jun 27, 2018

I ( n ) = 0 1 ( x n ( n + 1 ) 2 1 ) d x = 1 n + 1 ( n + 1 ) 2 1 = n \begin{aligned} I(n) = & \int_0^1 ( x^n(n+1)^2 -1) \,dx \\ = & \frac{1}{n+1} (n+1)^2 -1 \\ = & n \end{aligned} Now we want to use this information to evaluate 0 1 ( 16 x 3 1 ) d x \displaystyle \int_0^1 (16x^3 - 1)\,dx . First notice that 16 = ( 3 + 1 ) 2 16 = (3+1)^2 so what we have is 0 1 ( x 3 ( 3 + 1 ) 2 1 ) d x \displaystyle \int_0^1 (x^3(3+1)^2 -1) \,dx , meaning n = 3 I(3) = 3 n=3 \, \, \, \, \therefore \, \, \, \, \text{I(3)} = \boxed{3} .

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