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April 21st, 2018, 11:45 PM   #1
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Integral

What is the integral of the mathematical expression in the picture:
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 April 21st, 2018, 11:53 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra That's an inequality. You can't integrate an inequality. However, we can integrate the left hand expression of the inequality by a substitution of $u=x+1$. Alternatively, \begin{align*}\frac{x^2}{x+1} &= \frac{(x+1)^2 - 2x - 1}{x+1} \\ &= \frac{(x+1)^2 - 2(x+1) + 1}{x+1} \\ &= (x+1) - 2 + \frac1{x+1} \\ &= x - 1 +\frac1{x+1}\end{align*} Last edited by v8archie; April 22nd, 2018 at 12:02 AM.
April 21st, 2018, 11:58 PM   #3
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Quote:
 Originally Posted by shaharhada What is the integral of . . .
Did you mean “integral”? You posted in Algebra, not Calculus.

April 22nd, 2018, 12:05 AM   #4
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This is my solution.
Is it wrong?
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 April 22nd, 2018, 12:13 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 The fact that f(x)< 1 does NOT mean that $\displaystyle \int f(x)dx< 1$! For one thing, a indefinite integral like this always includes an arbitrary constant. The integral can be larger or less than 1 depending upon that constant. For a definite integral it is true that if $\displaystyle f(x)< 1$ then $\displaystyle \int_a^b f(x)dx< b- a$. Thanks from topsquark
April 22nd, 2018, 12:40 AM   #6
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And if the f(x) < 2, is it the same?
Can you give some examples for:
Quote:
 Originally Posted by Country Boy;593137 For a [b definite[/b] integral it is true that if $\displaystyle f(x)< 1$ then $\displaystyle \int_a^b f(x)dx< b- a$.

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