Comparing Areas of x^2 and x^3 in [0,1] Interval

[moham_87]

hi, I'm sorry I'm asking alot, but that time i need help

here the question says:
* Verify the inequality without evaluating the definite integrals:

(integration [x^2] from 0 to 1) >= (integration [x^3] from 0 to 1)

i can't solve that question without evaluating the integral
so how can i start it?

thank u a lot ...and any efforts will be appreciated

N.B. if anyone please could inform me how to write mathematical equations in a better form.

 

[Njorl]

I'll give you a hint. Can you show that for all points on the interval x^2>=x^3?

Njorl

 

[PrudensOptimus]

Originally posted by moham_87
hi, I'm sorry I'm asking alot, but that time i need help

here the question says:
* Verify the inequality without evaluating the definite integrals:

(integration [x^2] from 0 to 1) >= (integration [x^3] from 0 to 1)

i can't solve that question without evaluating the integral
so how can i start it?

thank u a lot ...and any efforts will be appreciated

N.B. if anyone please could inform me how to write mathematical equations in a better form.

take the dy/dx of both sides.

 

[Hurkyl]

Here's how you would write the equation:

<br /> \int_0^1 x^2 \, dx \geq \int_0^1 x^3 \, dx<br />

(click the image to see the source code)

 

[himanshu121]

You draw the graphs for x^2 & x^3
In the interval [0,1]
The area under the graph for x^2 is greater than x^3


\Rightarrow\int_0^1 x^2 \, dx \geq \int_0^1 x^3 \, dx