Question

help with integrals

Answer 1

the integral from [a,b]=[0,6] therefore the answer is (b)

Answer 2

for the first part a, it's 4*(-1/2 the area of a circle of radius 1)

Answer 3

hi andrea is that a circle or just like a sine wave

Answer 4

its an area from [0,2] + an area from [2,6] = area from [0,6] therefore ( b)

Answer 5

its a sine wave

Answer 6

nah looks like a circle to me

Answer 7

it is 2 circles from 0 to 1 it is a circle of radius 1, then from 2 to 6 it is another circle of radius 2 yeah

Answer 8

i'd bet the first one is \(-2\pi\)

Answer 9

yes

Answer 10

oh i thougth this question was old

Answer 11

did you get all the answers?

Answer 12

no the first one was corret

Answer 13

im not sure how you got that though

Answer 14

oh allow me to explain

Answer 15

you have half a circle, with radius 1. the whole circle would have area \(\pi r^2=\pi\) but you only have half the circle, so the area is \(\frac{\pi}{2}\) but of course it is below the x axis so it is negative and therfore the integral \[\int_0^2f(x)dx=-\frac{\pi}{2}\] and therefore
\[\int_0^24f(x)dx=4\int_0^2f(x)dx=4\times (-\frac{\pi}{2})=-2\pi\]

Answer 16

now
\[\int_2^6f(x)dx=\frac{1}{2}\pi (2)^2=2\pi\] i.e. again you have half a circle with radius 2 so area is half of \(4\pi\)

Answer 17

you were not asked for that integral i know but you were asked for
\[\int_0^64f(x)dx=4\int_0^2f(x)dx+4\int_2^6f(x)dx=-2\pi+8\pi=6\pi\]

Answer 18

similar reasoning solves the other ones

Answer 19

you got this? whole idea is to break it up in to pieces that you know the area of, and them piece them together. problem made more annoying by these stupid unnecessary constants like 4 and 5

Answer 20

i think i have it thank you

Answer 21

ok good. next one you are going to have to take one quarter of the circles and don't forget that area below the curve shows up negative