# Homework Help: Process for solving integration problems

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1. Nov 10, 2015

### Niaboc67

. The problem statement, all variables and given/known data
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:

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

3. 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: Nov 10, 2015
2. Nov 11, 2015

### Samy_A

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: Nov 11, 2015
3. Nov 11, 2015

### Ray Vickson

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 $-2x \leq x g(x) \leq 2x$?

4. Nov 12, 2015

### Niaboc67

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

5. Nov 12, 2015

### SammyS

Staff Emeritus
Right.

So the first line in the OP does have a typo.
$\ -2x\le xg(x)\le 2x\$​
$\ -2x\le g(x)\le 2x\$​