Process for solving integration problems

Click For Summary
The discussion focuses on solving integration problems involving continuous functions with specified bounds. For Q1, the correct interpretation of the bounds for g(x) is clarified, emphasizing that it should be -2 ≤ g(x) ≤ 2 rather than involving multiplication by x. The solutions provided for both Q1 and Q2 include calculations for lower and upper bounds of the integrals, with specific integrals evaluated. There are corrections noted regarding the assumptions made in the original attempts, particularly concerning the square root calculations and the bounds used. Overall, the thread highlights the importance of accurately interpreting function constraints in integration problems.
Niaboc67
Messages
249
Reaction score
3
. Homework Statement
Q1. Let f(x) be any continuous function that satisfies: $$-2x≤xg(x)≤2x$$ for $$0≤x≤1$$ Find the upper and lower bounds for:

$$\int_{0}^{1} \sqrt{1+g(x)+x^2}dx$$

Q2. Let h(x) be any continuous function that satisfies: $$-4≤h(x)≤x^2-4$$ for $$0≤x≤1$$ Find the upper and lower bounds for:

$$\int_{0}^{1}x^3h(x)dx$$

The Attempt at a Solution



Soltn for Q1 for lower bounds: $$\int_{0}^{1} \sqrt{1+g(x)+x^2}dx \\

= \int_{0}^{1} \sqrt{1-2x+x^2}dx \\

= \int_{0}^{1} \sqrt{(x-1)^2}dx \\

L_{f}(P) = \int_{0}^{1} (x-1)dx$$

Soltn for Q2 for upper bounds

$$\int_{0}^{1} \sqrt{1+2x+x^2}dx \\
= \int_{0}^{1} \sqrt{(x+1)^2}dx \\
U_{f}(P) = \int_{0}^{1} (x+1)dx$$Soltn for Q2 for lower bound: $$\int_{0}^{1}x^3h(x)dx \\

= \int_{0}^{1}x^3(-4)dx \\
L_{f}(P) = \int_{0}^{1}-4x^3dx$$

Soltn for Q2 for upper bound...

$$\int_{0}^{1}x^3(x^2-4)dx\\
U_{f}(P) = \int_{0}^{1}x^6-4x^3dx$$Thank you
 
Last edited by a moderator:
Physics news on Phys.org
You have to check your computations.

3 hints:
For Q1, you seem to assume that ##-2x≤g(x)≤2x##, but that is not what the question states.
When you compute a square root, you have to take the positive root.
##x^3x^2=x^5##
 
Last edited:
Niaboc67 said:
. Homework Statement
Q1. Let f(x) be any continuous function that satisfies: $$-2x≤xg(x)≤2x$$ for $$0≤x≤1$$ Find the upper and lower bounds for:

$$\int_{0}^{1} \sqrt{1+g(x)+x^2}dx$$

Q2. Let h(x) be any continuous function that satisfies: $$-4≤h(x)≤x^2-4$$ for $$0≤x≤1$$ Find the upper and lower bounds for:

$$\int_{0}^{1}x^3h(x)dx$$

The Attempt at a Solution



Soltn for Q1 for lower bounds: $$\int_{0}^{1} \sqrt{1+g(x)+x^2}dx \\

= \int_{0}^{1} \sqrt{1-2x+x^2}dx \\

= \int_{0}^{1} \sqrt{(x-1)^2}dx \\

L_{f}(P) = \int_{0}^{1} (x-1)dx$$

Soltn for Q2 for upper bounds

$$\int_{0}^{1} \sqrt{1+2x+x^2}dx \\
= \int_{0}^{1} \sqrt{(x+1)^2}dx \\
U_{f}(P) = \int_{0}^{1} (x+1)dx$$Soltn for Q2 for lower bound: $$\int_{0}^{1}x^3h(x)dx \\

= \int_{0}^{1}x^3(-4)dx \\
L_{f}(P) = \int_{0}^{1}-4x^3dx$$

Soltn for Q2 for upper bound...

$$\int_{0}^{1}x^3(x^2-4)dx\\
U_{f}(P) = \int_{0}^{1}x^6-4x^3dx$$Thank you

In Q1, can't you just say that ##-2 \leq g(x) \leq 2##? How does it help to multiply everything by ##x \geq 0 ## to get ##-2x \leq x g(x) \leq 2x##?
 
That was how the question was defined. It said: let g(x) be any continuous function that satisifes -2x≤g(x)≤2x for 0≤x≤1. Find the upper and lower bounds for ∫[from 0 to 1] √(1+g(x)+x^2)dx

Thanks
 
Niaboc67 said:
That was how the question was defined. It said: let g(x) be any continuous function that satisifes -2x≤g(x)≤2x for 0≤x≤1. Find the upper and lower bounds for ∫[from 0 to 1] √(1+g(x)+x^2)dx

Thanks
Right.

So the first line in the OP does have a typo.
You had:
##\ -2x\le xg(x)\le 2x\ ##​

It turns out it should be
##\ -2x\le g(x)\le 2x\ ##​

This makes more sense.
 
First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...