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Its DEFINITE integration between +1 and -1
(3x(squared) + 2x - 4) dx

2007-05-03 01:43:03 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

5 answers

(3x^(2) + 2x - 4) dx = 2x^3 + x^2 -4x +c
replace x=1
(2x^3 + x^2 -4x +c) = 2 + 1 - 4 + c
= -1 + c
=c - 1

replace x= -1
(2x^3 + x^2 -4x +c) = -2 + 1 + 4 + c
=3 + c

integral (3x^(2) + 2x - 4) dx = (c - 1) - (3 + c)
= - 4

2007-05-03 02:15:42 · answer #1 · answered by Tubby 5 · 0 1

First do the integration:
You get
x³ + x² -4x.
Now plug in the limits:
x = 1: 1 + 1 -4 = -2
x = -1: -1 + 1 + 4 = 4
Subtract to get the value of the integral = -6.

2007-05-03 05:31:10 · answer #2 · answered by steiner1745 7 · 0 0

3x^2 + 2x - 4 dx

Integrated-: x^3 + x^2 - 4x

Plug in 1 first....

1 + 1 - 4 = -2 (Result 1)

Now plug in -1....

-1 + 1 + 4 = 4 (Result 2)

Now subtract Result 2 from Result 1.....
-2 - 4 = -6

NB... A definate integral does not need the constant 'c' putting in, as the integral is 'definite'.

2007-05-03 01:50:53 · answer #3 · answered by Doctor Q 6 · 0 0

the integral is
x^3 +x^2 -4x +c

Now evaluate it (plug in) at +1 and -1
+1:
1+1-4+c =-2+c

-1:
-1+1+4+c= 4+c

Now subtract the first minus the second:
(-2+c)-(4+c)
The cs cancel and we get
-6

2007-05-03 01:48:15 · answer #4 · answered by Anonymous · 0 0

positioned sqrt (a million+3x)=t =>a million+3x=t^2 => 3dx=2tdt=> dx=2/3tdt top reduce : x=5 ,so t=sqrt (a million+15)=4 . decrease reduce : x=0,so t=a million. for this reason very last answer is (2/3)crucial tdt/t = (2/3)crucial dt (from a million to 4) = (2/3) (4-a million)=2 . merely examine the calculations.

2016-12-05 07:00:09 · answer #5 · answered by abila 4 · 0 0

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