Integration help (2 Viewers)

z3bra · Jul 19, 2017

Can someone please help me with q8?

Sent from my iPhone using Tapatalk

Drongoski · Jul 19, 2017

I think there is a typo in the question. The 2nd integral has to be:

$\int ^7 _0 [ f(x-3) + 2] dx$

Answer = 14

PS: Maybe not a typo; maybe the examiner who copied or set this question could not tell the error

1729 · Jul 19, 2017

z3bra said:
Can someone please help me with q8?

Sent from my iPhone using Tapatalk

$$\noindent Is the question written correctly? Because why wasn't it just written $f(x-1)$? \\\\ If this is the case then $\int_0^4 f(x)dx = \int_3^7 f(x-3)dx$ as the graph along with the bounds have all been shifted 3 units to the right. You can then split up the integral and work out $\int_3^72dx$ and sum the results. \\\\ Note: the above concerns the integral $\int_3^7\left[f(x-3) + 2\left]dx = 6 + 8 = 14$

z3bra · Jul 19, 2017

Drongoski said:
I think there is a typo in the question.

Why do u think the typo is? With the limits or where the brackets are.

Sent from my iPhone using Tapatalk

z3bra · Jul 19, 2017

1729 said:
$$\noindent Is the question written correctly? Because why wasn't it just written $f(x-1)$? \\\\ If this is the case then $\int_0^4 f(x)dx = \int_3^7 f(x-3)dx$ as the graph along with the bounds have all been shifted 3 units to the right. You can then split up the integral and work out $\int_3^72dx$ and sum the results. \\\\ Note: the above concerns the integral $\int_3^7\left[f(x-3) + 2\left]dx$

Right. Thanks ya I think your right. There's placement of brackets at wrong. Ur assumption is right and I got the right answer c. Thanks.

Sent from my iPhone using Tapatalk

Drongoski · Jul 19, 2017

z3bra said:
Why do u think the typo is? With the limits or where the brackets are.

Sent from my iPhone using Tapatalk

In the form given you won't be able to find the answer.

Drongoski · Jul 19, 2017

$$Say: $ \int f(x) dx = F(x) + C \implies F(4) - F(0) = 6\\ \\ \int _3 ^7[f(x-3) + 2] dx = \int ^7 _3 f(x-3) d(x-3) + \int ^7 _0 2dx = \left [ F(x-3) + 2x \right ]^7 _3 = (F(4) - F(0) ) + 8 = 6 + 8 = 14$

Note that we only know F(4) - F(0) = 6. We don't even know F(4) = ? and F(0) = ? much less F(a) for other values of 'a'. Hope this makes things clearer, although many would wonder about d(x-3).

Integration help (2 Viewers)

z3bra

New Member

Drongoski

Well-Known Member

1729

Active Member

z3bra

New Member

z3bra

New Member

Drongoski

Well-Known Member

Drongoski

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)