What is the correct solution for this integral?

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In summary, the student attempted to solve the given integral by substituting u= ae^theta-b and e^theta= (u+b)/a. However, they did not properly substitute and got an incorrect answer. The correct solution involves changing dtheta to du and results in the integral equal to 2ln(ae^theta-b) - theta + C.
  • #1
alba_ei
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Homework Statement



[tex]\int \frac{ae^\theta+b}{ae^\theta-b} \, d\theta[/tex]

The Attempt at a Solution



i took [tex]u = ae^\theta-b[/tex] so [tex]e^\theta = \frac{u + b}{a}[/tex] then i substituded back into the integral and iget this

[tex]\int \frac{u + b + b}{u} \, du[/tex]

[tex]\int du +\int \frac{2b}{u} \, du[/tex]

[tex]= u \du + 2b \ln u +C[/tex]

[tex]= u + 2b \ln u +C[/tex]

[tex]= ae^\theta-b + 2b\ln (ae^\theta-b) [/tex]

but the answer of the book is
[tex]\int \frac{ae^\theta+b}{ae^\theta-b} \, d\theta = 2\ln (ae^\theta-b) - \theta + C [/tex]
what did i do wrong?
 
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  • #2
You didn't subsitute properly. You have to change dtheta too.
 
  • #3
write d0 as what it should equal to du
for example if u=x^2
du=2*xdx
 

FAQ: What is the correct solution for this integral?

What is an integral?

An integral is a mathematical concept used to calculate the area under a curve in a given interval. It is also known as the antiderivative, as it is the inverse operation of differentiation.

Why are integrals important?

Integrals are important in various fields, including physics, engineering, and economics. They allow us to calculate important quantities such as displacement, velocity, acceleration, and work done.

How do you solve an integral?

Integrals can be solved using various techniques such as substitution, integration by parts, and partial fractions. The method used depends on the form of the integral and the functions involved.

What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration, and the result is a single numerical value. On the other hand, an indefinite integral does not have limits of integration, and the result is a function.

Can integrals be used for other shapes besides curves?

Yes, integrals can be used to find the volume and surface area of 3-dimensional shapes, such as spheres, cylinders, and cones. They can also be used to find the length of a curve and the area of a region in the xy-plane.

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