My fifth integral problem
∫ − 3 1 x 2 + 4 x 2 + 1 6 x 2 + 6 4 x 2 + ⋯ d x = ?
Clarification : The coefficient of x 2 in each successive nested radical is in the form of 4 n − 1 , where n is a positive integer.
The answer is 9.
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2 solutions
Your graph is wrong. the vertex needs to be at x=-1 instead of x= 1 however the area will be 4
I agree with this solution, why has been the answer changed?why the solution is 9, now?
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I also couldn't know where I went wrong...... See now only 2 percent got this question right.....
∣ x − 1 ∣ = x 2 − 2 x + 1 = x 2 + 4 x 2 − 4 x + 1 = x 2 + 4 x 2 + 1 6 x 2 − 8 x + 1 = x 2 + 4 x 2 + 1 6 x 2 + 6 4 x 2 − 1 6 x + 1 = … How do we tell the difference?
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Going from step 1 to 2 in your solution you make an assumption that 1>2x that's why you have written (1-2x)= 4 x 2 − 4 x + 1 and hence for following steps you have to ensure 4x<1, 16x<1,.. which is not the case since we are integrating from -3 to 1.
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I see now. Thank you!
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@John Frank – Can you find any flaw in this approach ?? Why is the answer been edited to 9??
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@Rishabh Jain – I think you should refer to Hobart Pao's updated solution.
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@John Frank – I have seen that but couldn't understand where things went wrong... But OK .... I'll delete my solution after some time....
May you tell me how you knew that it was the absolute value of x + 1 ? I had not thought of the absolute value sign and so I was very confused with this question.
UPDATED SOLUTION Inspiration from Amazing Nested Radical 3 The point of this problem is to consider the following: ∣ x + 1 ∣ can be rewritten as ( x + 1 ) 2
= x 2 + 2 x + 1
= x 2 + ( 2 x + 1 ) 2
= x 2 + 4 x 2 + 4 x + 1
= x 2 + 4 x 2 + ( 4 x + 1 ) 2
= x 2 + 4 x 2 + 1 6 x 2 + 8 x + 1
and so on.
However, because we have x 2 all over the place, then the function becomes an even function. If you have a negative value of x , you get the positive value of x for f ( x ) . So,
∫ − 3 1 f ( x ) d x = ∫ − 3 0 f ( x ) d x + ∫ 0 1 f ( x ) d x = ∫ − 3 0 ( − x + 1 ) d x + ∫ 0 1 ( x + 1 ) d x = 9 .
Can you tell what's the flaw in previous approach...???
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Because ∣ x + 1 ∣ is not an even function, but since we have x 2 all over the place in the integral, we have to make ∣ x + 1 ∣ even by taking absolute value of the x inside.
Did it the same way!
This problem is a beautiful modification and extension of 3 . Other inspirations might be 2 and 1 .Note: ∣ x + 1 ∣ = ( x + 1 ) 2 = x 2 + 2 x + 1 = x 2 + ( 2 x + 1 ) 2 = x 2 + 4 x 2 + 4 x + 1 = x 2 + 4 x 2 + ( 4 x + 1 ) 2 = x 2 + 4 x 2 + 1 6 x 2 + 8 x + 1 = ⋯ which is what required in the question hence ∫ − 3 1 ∣ x + 1 ∣ = 4 . . . . . see figure