-
What is the main difference between a proper and an improper integral?
-
Why do we need limits when defining an improper integral?
-
Can you give a physical example where an improper integral naturally appears (e.g., in electrostatics or thermodynamics)?
-
Do all improper integrals with infinity as a bound diverge? Explain your reasoning.
-
Optional: Share one example of an improper integral you have encountered and explain whether it converges or diverges.
1. Proper integrals are finite and well-behaved; improper integrals involve infinity or blow-ups.
2. We use limits because the integral cannot be evaluated directly at infinity or at singular points.
3. Example: calculating the field of an infinite line charge in electrostatics.
4. No — some improper integrals converge if the function decreases fast enough; others diverge.
2. We use limits because the integral cannot be evaluated directly at infinity or at singular points.
3. Example: calculating the field of an infinite line charge in electrostatics.
4. No — some improper integrals converge if the function decreases fast enough; others diverge.
Prosper intégral are finit and thé fonction IS continuons
impropre intégral are Infinite
2-we use limite because WE cannot evaluate a fonction directly at Infinite
3-
4-no
5-intgral e^-x DX
impropre intégral are Infinite
2-we use limite because WE cannot evaluate a fonction directly at Infinite
3-
4-no
5-intgral e^-x DX
1. Main Difference A proper integral is calculated over a bounded interval with a bounded function, whereas an improper integral deals with either unbounded intervals (such as infinity) or unbounded functions.
2. The integral cannot be directly calculated at a point of discontinuity or at infinity, so we use a limit to study its behavior.
3. It appears in physics when calculating the electric potential of a very long (infinite) charged wire.
4. No some converge if the function decreases rapidly enough (such as the integral of 1/x²).
5. Example: The integral of 1/√x from 0 to 1 is an improper integral (a singularity at zero)
2. The integral cannot be directly calculated at a point of discontinuity or at infinity, so we use a limit to study its behavior.
3. It appears in physics when calculating the electric potential of a very long (infinite) charged wire.
4. No some converge if the function decreases rapidly enough (such as the integral of 1/x²).
5. Example: The integral of 1/√x from 0 to 1 is an improper integral (a singularity at zero)
1:Proper integral:
The integrand is finite and continuous on a finite interval
Improper integral:
At least one of the following occurs:
The interval is infinite (e.g.
The function is unbounded at some point in the interval
These integrals are defined using limits.
2:Because the integral cannot be evaluated directly:
The upper or lower bound may be infinite
Or the function may blow up at some point
Limits allow us to:
Replace the problematic bound or point
3:Electrostatics:
The electric potential U due to a point charge is computed by integrating the electric field from a point to infinity
Thermodynamics:
The partition function often involves integrals over all energies
4:No.
Some converge, others diverge, depending on how fast the function decreases.
Examples:
Convergent:
Divergent:
So infinity alone does not imply divergence.
5:
السلام عليكم ورحمة الله بركاته اختي في الإجابة رقم 4 لم تظهر الامثله ربما حدث خطأ ما
اجل
لقد حاولة الاجابة على شكل ملف لكن الحجم كان اكبر
🥰شكرا على الملاحظة
الاجابة هي :
∫(from 1 to ∞) (1/𝑥²) 𝑑𝑥 converges,
while ∫(from 1 to ∞) (1/𝑥) 𝑑𝑥 diverges.
Oki
هل يمكن ان تعطيني مثال اخر على التكامل الغير صحيح؟
السوال 5
وبخصوص سؤال رقم 5 الاجابة كالآتي :
Example: ∫(from 0 to ∞) e^{-x} \, dx
· Reason it is improper: The interval is unbounded ( [0, \infty) ).
· Convergence check: Using the limit:
\lim_{b \to \infty} \int_{0}^{b} e^{-x} \, dx = \lim_{b \to \infty} (1 - e^{-b}) = 1
· Result: The integral converges and its value is 1.
Example: ∫(from 0 to ∞) e^{-x} \, dx
· Reason it is improper: The interval is unbounded ( [0, \infty) ).
· Convergence check: Using the limit:
\lim_{b \to \infty} \int_{0}^{b} e^{-x} \, dx = \lim_{b \to \infty} (1 - e^{-b}) = 1
· Result: The integral converges and its value is 1.
1/Proper Integral: Has finite limits and a bounded (continuous) function.
Improper Integral: Has at least one infinite limit (\infty or -\infty) or a function that becomes infinite (vertical asymptote) within the interval.
2/We use limits because we cannot evaluate a function at infinity or at a point where it is undefined.
Improper Integral: Has at least one infinite limit (\infty or -\infty) or a function that becomes infinite (vertical asymptote) within the interval.
2/We use limits because we cannot evaluate a function at infinity or at a point where it is undefined.
3/Calculating the energy required to move an object from a planet's surface to infinity involves an improper integral.
4/No. An improper integral with an infinite bound converges if the function approaches zero fast enough.
4/No. An improper integral with an infinite bound converges if the function approaches zero fast enough.
رداً على Souad Ayadi
Understanding and Interpreting Improper Integrals
بواسطة - MAROUA AHMED ABDELMALEK
1. Proper Integrals
A proper integral is defined when the limits of integration are finite and the function remains continuous throughout the interval.
An integral is called improper if at least one of the following conditions is satisfied:
One or both integration limits are infinite.
The integrand is undefined or becomes unbounded at some point within the interval.
3. Definition Using Limits
Improper integrals are evaluated by expressing them as limits.
This approach allows us to determine whether the integral converges or diverges.
4. Physical Applications
Improper integrals frequently appear in physics.
In electrostatics, they are used to compute electric fields generated by charge distributions extending to infinity.
In thermodynamics, they help calculate the total energy of systems occupying unbounded regions of space.
5. Convergence and Divergence of Improper Integrals
An improper integral with infinite limits does not always diverge.
Its behavior depends on how the integrand behaves as � becomes large.
If the function decreases sufficiently fast, the integral converges.
If the function decreases too slowly or remains large, the integral diverges.