Question

Need help in these questions (in the picture) Thanks!

Answer 1

for #21, you may figure out the answer without actually evaluating the integral

Answer 2

definite integral represents the area under the curve in the interval over which you're integrating

Answer 3

Look at the integrand \(\dfrac{2}{\sqrt{9-4x^2}}\)


Over the given interval \((0,1)\),
notice that the minimum value is \(\dfrac{2}{\sqrt{9}} \)
and the maximum value is \(\dfrac{2}{\sqrt{5}} \)

Answer 4

that is, the integral must evaluate to some number between \(0.67\) and \(0.89\)

look at the options, there is only one option falling in that interval

Answer 5

hey i got kinda lost on the minimum and maximum value thing. haha. sorry i really need a refresh on this topics

Answer 6

but if it's possible how do i get it with integration, because i think they check the solutions in this exams.

Answer 7

is this an exam ?

Answer 8

no, it's a reviewer, the exam is on sunday. haha. what i meant was they need solutions for the upcoming exam. :)

Answer 9

sorry for the confusion. :)

Answer 10

okay good :)
if you want, i can explain that maximum and minimum thingy... it won't take much time

Answer 11

yes please, i really got interested on that one. haha

Answer 12

it is indeed a very interesting way to approximate the value of a definite integral

Answer 13

how good are you at interpreting the definite integral as area under the curve ?

Answer 14

not really good to be honest.