Evaluate the following integrals. int(dx)/(sqrt((x-1)(3-x)))

Yulia 2020-12-02 Answered

Evaluate the following integrals.

\int \frac{d x}{\sqrt{(x - 1) (3 - x)}}

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Expert Answer

wheezym

Answered 2020-12-03 Author has 103 answers

Step 1: Given that
Completing the square Evaluate the following integrals.

\int \frac{d x}{\sqrt{(x - 1) (3 - x)}}

Step 2: Formula Used

\int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = \sin^{- 1} (\frac{x}{a}) + C

Step 3:Solve
We have,

\int \frac{d x}{\sqrt{(x - 1) (3 - x)}} = \int \frac{d x}{\sqrt{3 x - x^{2} - 3 + x}}

= \int \frac{d x}{\sqrt{- x^{2} + 4 x - 3}}

= \int \frac{d x}{\sqrt{- (x^{2} - 4 x + 3)}}

= \int \frac{d x}{\sqrt{- (x^{2} - 4 x + 2^{2} - 2^{2} + 3)}}

= \int \frac{d x}{\sqrt{- {(x - 2)}^{2} - 4 + 3}}

= \int \frac{d x}{\sqrt{- ({(x - 2)}^{2}) - 1}}

= \int \frac{d x}{\sqrt{1 - {(x - 2)}^{2}}}

= \sin^{- 1} (x - 2) + C

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New questions

If h has positive derivative and

φ

is continuous and positive. Where is increasing and decreasing f
The problem goes specifically like this:
If h is differentiable and has positive derivative that pass through (0,0), and

φ

is continuous and positive. If:

f (x) = h (\int_{0}^{\frac{x^{4}}{4} - \frac{x^{2}}{2}} φ (t) d t) .

Find the intervals where f is decreasing and increasing, maxima and minima.
My try was this:
The derivative of f is given by the chain rule:

f^{'} (x) = h^{'} (\int_{0}^{\frac{x^{4}}{4} - \frac{x^{2}}{2}} φ (t) d t) φ (\frac{x^{4}}{4} - \frac{x^{2}}{2})

We need to analyze where is positive and negative. So I solved the inequalities:

\frac{x^{4}}{4} - \frac{x^{2}}{2} > 0 \land \frac{x^{4}}{4} - \frac{x^{2}}{2} < 0

(- \infty, - \sqrt 2) \cup (\sqrt 2, \infty)

for the first case and

(- \sqrt 2, \sqrt 2)

for the second one. Then (not sure of this part)

h^{'} (\int_{0}^{\frac{x^{4}}{4} - \frac{x^{2}}{2}} φ (t) d t) > 0

φ (\frac{x^{4}}{4} - \frac{x^{2}}{2}) > 0

x \in (- \infty, - \sqrt 2) \cup (\sqrt 2, \infty) .

Also if both h′ and

φ

are negative the product is positive, that's for

x \in (- \sqrt 2, \sqrt 2)

.
The case of the product being negative implies:

x \in [(- \infty, - \sqrt 2) \cup (\sqrt 2, \infty)] \cap (- \sqrt 2, \sqrt 2) = [(- \infty, - \sqrt 2) \cap (- \sqrt 2, \sqrt 2)] \cup [(\sqrt 2, \infty) \cap (- \sqrt 2, \sqrt 2)] = \emptyset .

So the function is increasing in

(- \infty, - \sqrt 2), (- \sqrt 2, \sqrt 2), (\sqrt 2, \infty)

. So the function does not have maximum or minimum. Not sure of this but what do you think?

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.
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and told that routine computations will show that

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x (1) = π

.
I must determine a differential equation for

x (t)

x^{'} (t) = f (t, x)

.
I must then use

f

π

.
I will use Euler's Method to find minimum number of iterations (

n

) needed to yield the approximation

π \approx 3.1400

n - 1

iterations is strictly less than 3.1400.
How do I find

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