2 integrals - - (within next 30 mins if possible)

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In summary: I don't really understand what you are asking for a summary of. Are you looking for a summary of the solution process or of the conversation between the two individuals? Please provide more context so I can provide an accurate summary.
  • #1
Mattofix
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2 integrals - URGENT - (within next 30 mins if possible)

Homework Statement



[tex]\int_0^{b}\frac{dx}{x+\sqrt x}[/tex]

[tex]\int[/tex][tex]^{b}_{}[/tex] (x + [tex]\sqrt{x})^{-1}[/tex]dx
[tex]_{ 1}[/tex]

The Attempt at a Solution



For the first i have tried x=t^2 but ended up with something i can't integrate (2t/(t^2 + t) dt

for the second i tried u= 1 + x^3 but that did'nt help -
 
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  • #2
Hint:
[tex]\frac{2t}{t^2 + t} = \frac{2t+1}{t^2 + t} - \frac{1}{t^2 + t}[/tex]

This screams substitution for the first term and partial fractions for the second.

Your second integrand is exactly like your first one. Am I missing something?
 
  • #3
ahhhh- yeah the second is

[tex]\int[/tex][tex]^{\infty}_{}[/tex] (1 + x[tex]^{3}[/tex])[tex]^{-1/2}[/tex] dx
[tex]_{ 1}[/tex]
 
  • #4
try the substitution

y=sqrt[x]

...oh, right... that is what you tried... then factor out a y from numerator and denominator...
 
  • #5
cool - cheers
 
  • #6
[tex]\int_0^{b}\frac{dx}{x+\sqrt x}[/tex]

This thing works out nicely if you factor out [tex]\sqrt{x}[/tex], giving you

[tex]\sqrt{x}\left(\sqrt{x} + 1\right)[/tex]

Then let u = [tex]\left(\sqrt{x} + 1\right)[/tex] and you're almost done.
 
  • #8
yeah - i got the x= t^2 but i could't integrate (2t/(t^2 + t) dt

thanks though
 
  • #9
Mattofix said:
yeah - i got the x= t^2 but i could't integrate (2t/(t^2 + t) dt

thanks though
i also simplified it in there, now you will end up with 2 ln(1+t) on the interval sqrt b, and 0
 

Related to 2 integrals - - (within next 30 mins if possible)

1. What are integrals?

Integrals are mathematical tools used to find the area under a curve or the accumulation of a quantity over an interval. They are an essential concept in calculus and are used in many fields of science, including physics, engineering, and economics.

2. What is the difference between a definite and indefinite integral?

A definite integral has specific limits of integration and gives a numerical value, while an indefinite integral does not have limits and gives a general function as a solution. Definite integrals are used to find precise values, while indefinite integrals are used to find antiderivatives.

3. How do you solve a double integral?

A double integral is solved by first integrating with respect to one variable and then integrating the resulting function with respect to the other variable. This process can be repeated for triple or higher order integrals. It is essential to correctly set up the limits of integration for each variable.

4. What is the fundamental theorem of calculus?

The fundamental theorem of calculus states that the definite integral of a function f(x) can be evaluated by finding the antiderivative of f(x) and evaluating it at the limits of integration. It provides a connection between differentiation and integration and is a fundamental concept in calculus.

5. How are integrals used in real life?

Integrals have many real-life applications, such as calculating the area under a velocity-time graph to find the displacement of an object, finding the work done by a force, or determining the average value of a function. They are also used in fields like economics to model supply and demand curves and in engineering to calculate volumes and surface areas of 3D objects.

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