Question

Use the properties of integrals to verify the inequality.

Answer 1

For the first problem, note that since x goes from 0 to 1:\[0<\cos(x)\le 1\]This tells us that:\[x^2\cos(x)\le x^2 \]for that interval. Now we take the integral of both sides:\[x^2\cos(x)\le x^2 \Longrightarrow \int\limits_0^1x^2 \cos(x)dx\le \int\limits_0^1x^2 dx\]The right integral is easier to solve.

Answer 2

how did you get from \[0<x<1 \to 0<\cos(x)<1\]

Answer 3

i plug cos into 0 and 1 and i get 1<cosx<0.54

Answer 4

i understand the steps after that but im confused on 0<x<1→0<cos(x)<1

Answer 5

did you look at the cos(x) graph to see that 0<cos(x)<1 from [0,1]

Answer 6

is that how

Answer 7

oh sry, i wasnt looking at the thread, my bad >.< that is right, when you plug in 0 you get 1, and when you plug in 1 you get .54

What i was trying to say was that for x between 0 and 1, cos is always positive, and its value is less than or equal to 1. you can disregard the 0<cosx<1 thing.

Answer 8

what mistakes i made on 41
\[pi/4\le x \le pi/2 \]
\[\sin \le 1\]
\[\sin(x)/x \le x^-1\]
\[\int\limits_{pi/4}^{pi/2}\sin(x)/x \le \int\limits_{pi/4}^{pi/2} x^-1\]
\[\int\limits\limits_{pi/4}^{pi/2}\sin(x)/x \le 0\]

Answer 9

i mean 42