Question

Integration @ganeshie8

Answer 1

\[Given~\int\limits\limits_{-1}^{2}[f(x)+1]~dx=9~and~\int\limits\limits_{2}^{5}f(x)~dx=16,find\]

Answer 2

\[a)\int\limits_{-1}^{2}3f(x)~dx\]

Answer 3

\[b)\int\limits_{-1}^{5}[2f(x)-1]dx\]

Answer 4

integral has below properties :

\[\int \color{red}{k}f(x) \, dx = \color{red}{k}\int f(x)\,dx\]

\[\int [f(x)+g(x)] \, dx = \int f(x)\,dx+\int g(x)\,dx\]

Answer 5

We're given : \[\int\limits\limits_{-1}^{2}[f(x)+1]~dx=9\]

using the previously listed integral properties, can you find the value of \(\int\limits_{-1}^2f(x)\, dx\) ?

Answer 6

\[\int\limits\limits_{-1}^{2}f(x)+\int\limits_{-1}^{2}1~dx\]

Answer 7

Yes, keep going...

Answer 8

don't forget the dx in first integral

Answer 9

\[\int\limits\limits_{-1}^{2}f(x)\,\color{red}{dx}+\int\limits_{-1}^{2}1~dx\]

Answer 10

\[[\frac{ x^2 }{ 2 }]_{-1}^{2}+[x]_{-1}^{2}=9\]

Answer 11

nope

Answer 12

we don't know what f(x) is
so we cannot integrate it directly

Answer 13

\[\int\limits_{-1}^{2}f(x)~dx=6\]

Answer 14

We're given :
\(\int\limits\limits_{-1}^{2}[f(x)+1]~dx=9\)

\(\int\limits\limits_{-1}^{2}f(x)~dx+\int\limits\limits_{-1}^{2}1~dx=9\)

\(\int\limits\limits_{-1}^{2}f(x)~dx = 9 - \int\limits\limits_{-1}^{2}1~dx\)

\(\int\limits\limits_{-1}^{2}f(x)~dx = 9 - 3 = 6\)

Answer 15

So, we have two things :

\[\int\limits_{-1}^{2}f(x)~dx=6\tag{1}\]
\[\int\limits_{2}^{5}f(x)~dx=16\tag{2}\]

we're ready to answer the questions

Answer 16

\[a)\int\limits_{-1}^{2}3f(x)~dx\]

do you want to try working this ?
you just need to use one of the previously listed integral properties

Answer 17

okay,i will try

Answer 18

\[3\int\limits_{-1}^{2}f(x)~dx=3(6)=18\]

Answer 19

Perfect !

Answer 20

for question b)\[\int\limits_{-1}^{5}[2x-1]~dx=2\int\limits_{-1}^{5}[f(x)-\frac{ 1 }{ 2 }]dx\] is it correct?

Answer 21

\[\int\limits_{-1}^{5}f(x)=22\]

Answer 22

that is correct, but i feel splitting like below looks better :

\[\int\limits_{-1}^{5}[2f(x)-1]~dx=\int\limits_{-1}^{5}2f(x)\,dx-\int\limits_{-1}^{5}1\,dx\\~\\~\\
=2\int\limits_{-1}^{5}f(x)\,dx-\int\limits_{-1}^{5}1\,dx\\~\\~\\\]

Answer 23

\[\int\limits_{-1}^{5}f(x)=22\]

how did u get this ?

Answer 24

i think i did wrong :\
6+16=22

Answer 25

that is correct! i just wanted to know how you worked it, thats all :)

Answer 26

oh i c... :D

Answer 27

As you can see, definite integrals just add up like areas :
\[\int\limits_a^{\color{red}{c}}f(x)\,dx + \int\limits_{\color{red}{c}}^b f(x)\, dx = \int\limits_a^bf(x)\, dx\]

Answer 28

(Area under curve between a to c ) + (Area under curve between c to b ) = (Area under curve between a to b )

Answer 29

\[\int\limits_{-1}^{5}[2f(x)-1]=44-6=38\]

Answer 30

Thank you @ganeshie8 :)

Answer 31

Looks good! remember, this is calculus, knowing how to work some particular problem is not enough... you need to have very good understanding of graphs of functions to really enjoy calculus

Answer 32

okay, thnx @ganeshie8