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Natali5045456 [20]
3 years ago
6

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Mathematics
2 answers:
maw [93]3 years ago
4 0

Answer:

Convergent; \frac{1}{32}.

Step-by-step explanation:

We have been given an integral as \int _3^{\infty }\:\:\frac{1}{\left(x+1\right)^3}\:dx. We are asked to determine whether our given integral converges or diverges.

Let us integrate our given integral by u substitution as:

\int _3^{\infty }\:\:\frac{1}{\left(u)^3}\:dx

\int _3^{\infty }\:\:u^{-3}\:dx

\int _3^{\infty }\:\:u^{-3}\:dx=\frac{u^{-3+1}}{-3+1}

\int _3^{\infty }\:\:u^{-3}\:dx=\frac{u^{-2}}{-3+1}

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

Now, we will compute the boundaries.

-\frac{1}{2(\infty+1)^2}=-\frac{1}{\infty ^2}=0

-\frac{1}{2(3+1)^2}=-\frac{1}{2(4^2}=-\frac{1}{2*16}=-\frac{1}{32}

Our definite integral would be 0-(-\frac{1}{32})=\frac{1}{32}

Therefore, our given integral is convergent and its value is \frac{1}{32}.

Anika [276]3 years ago
3 0

Answer:

I=\dfrac{1}{32}

Step-by-step explanation:

Given,

Improper Integral I is given as  

I=\int^{\infty}_{3}\frac{1}{(x+1)^3}dx

integration of

\dfrac{1}{(x+1)^2}

now 

\int \dfrac{1}{(x+1)^3}dx = \dfrac{(x+1)^{-3+1}}{-3+1}

\int \dfrac{1}{(x+1)^3}dx =\dfrac{1}{-2(x+1)^2}

I=\left [\dfrac{1}{-2(x+1)^2}\right]^{\infty}_3

substituting value  

I=-\dfrac{1}{2}\left [ \dfrac{1}{\infty^2}-\dfrac{1}{4^2}\right ]

I=-\dfrac{1}{2}\left [-\dfrac{1}{16}\right ]

I=\dfrac{1}{32}

so the value of integral converges at \frac{1}{32}

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

Bases od isosceles triangles are congruent,so

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All angles in a triangle add to 180

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Using a system of equations, it is found that since 20 child bikes and 6 adult bikes would require more testing than the allocated time, it is not possible to build this amount.

<h3>What is a system of equations?</h3>

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are:

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Each child bike requires 4 hours to build, as do each adult bike. The company has 100 hours of testing, hence:

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c + a = 25.

With 20 child bikes and 6 adult bikes in a week, we have that c = 20, a = 26, hence:

c + a = 26

20 child bikes and 6 adult bikes would require more testing than the allocated time, it is not possible to build this amount.

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For the first problem the answer is -5/2

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Step 1: Simplify both sides of the equation.
2x−7+10=−2
2x+−7+10=−2
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(Combine Like Terms)
2x+3=−2

Step 2: Subtract 3 from both sides.
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Answer:
x= -5/2
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