Chapter 5 | Integration
607
x dt t , using properties of the definite
[T] Compute the right endpoint estimates
390.
definition ln( x ) = ∫ 1
⌠ ⌡ −3 5
1 2 2 π
e −( x −1) 2 /8 .
R 50 and R 100 of
integral and making no further assumptions.
383. Use the identity ln( x ) = ∫ 1 x dt
t to derive the identity
⎛ ⎝ 1 x
⎞ ⎠ =−ln x .
ln
384. Use a change of variable in the integral ∫ 1 xy 1
t dt to
show that ln xy = ln x +ln y for x , y >0.
385. Use the identity ln x = ∫ 1 x dt x to show that ln( x ) is an increasing function of x on [0, ∞), and use the previous exercises to show that the range of ln( x ) is (−∞, ∞). Without any further assumptions, conclude that ln( x ) has an inverse function defined on (−∞, ∞). 386. Pretend, for the moment, that we do not know that e x is the inverse function of ln( x ), but keep in mind that ln( x ) has an inverse function defined on (−∞, ∞). Call it E . Use the identity ln xy = ln x +ln y to deduce that E ( a + b ) = E ( a ) E ( b ) for any real numbers a , b . 387. Pretend, for the moment, that we do not know that e x is the inverse function of ln x , but keep in mind that ln x has an inverse function defined on (−∞, ∞). Call it E . Show that E '( t ) = E ( t ). 388. The sine integral, defined as S ( x ) = ∫ 0 x sin t t dt is an important quantity in engineering. Although it does not have a simple closed formula, it is possible to estimate its behavior for large x . Show that for k ≥1, | S (2 πk )− S ⎛ ⎝ 2 π ( k +1) ⎞ ⎠ | ≤ 1 k (2 k +1) π . ( Hint : sin( t + π ) =−sin t ) 389. [T] The normal distribution in probability is given by p ( x ) = 1 σ 2 π e −( x − μ ) 2 /2 σ 2 , where σ is the standard deviation and μ is the average. The standard normal distribution in probability, p s , corresponds to μ =0and σ =1. Compute the right endpoint estimates R 10 and R 100 of ⌠ ⌡ −1 1 1 2 π e − x 2/2 dx .
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