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nevsk [136]
3 years ago
14

Evaluate the integral: ∫dx/(sqrt((x^2)+16))

Mathematics
1 answer:
Lorico [155]3 years ago
7 0
Let x=4tanβ, then dx = 4sec²βdβ.  Now

    ∫(4sec²β)/(4secβ)dβ
=  ∫secβdβ
=  ln |tanβ - secβ| + c
=  ln |(x/4) - (√(x²+16)/4| + C
=  ln |x/√(x²+16)
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Answer:

0.5<2-√2<0.6

Step-by-step explanation:

The original inequality states that 1.4<√2<1.5

For the second inequality, you can think of 2-√2 as 2+(-√2).

Because of the "properties of inequalities", we know that when a positive inequality is being turned into a negative, the numbers need to swap and become negative. So, the original inequality becomes -1.5<-√2<-1.4. (Notice how the √2 becomes negative, too). This makes sense because -1.5 is less than -1.4.

Using our new inequality, we can solve the problem. Instead of 2+(-√2), we are going to switch "-√2" with both possibilities of -1.5 and -1.6. For -1.5, we would get 2+(-1.5), or 0.5. For -1.4, we would get 2+(-1.4), or 0.6.

Now, we insert the new numbers into the equation _<2-√2<_. The 0.5 would take the original equation's "1.4" place, and 0.6 would take 1.5's. In the end, you'd get 0.5<2-√2<0.6. All possible values of 2-√2 would be between 0.5 and 0.6.

Hope this helped!

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