Process for solving integration problems

Niaboc67
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. 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
 
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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##
 
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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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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