Question

evaluate ∫upper bound (3) lower bound (-2) (4t^3+6) dt

Answer 1

you just need the x^n formula of integration

\(\Large \int x^n dx = \dfrac{x^{n+1}}{n+1}+c\)

Answer 2

for the first term
4 is constant
for t^3, n = 3

for the 2nd term
6 is constant,
t^0
so,n=0

Answer 3

what i dont get is why the only possible answers are :

-103

-95

95

103

Answer 4

could you help me ??

Answer 5

thats because after integration, we plug in the upper and lower limits...


is this your first integration question ??

if not, have you tried to start to solve this one ?

Answer 6

so wait, do you plug the numbers in ... (x^(3+1)/(3+1)) + 4 = x^4/4 + 4

Answer 7

use 't'

so for 4t^3
you get
\(\Large 4\dfrac{t^4}{4}\)

got this ?

how about \(6=6t^0\) ?

Answer 8

i dont get it .. how do you use t to get that equation?

Answer 9

the variable in your question is 't'

so,
\(\Large \int t^n dt = \dfrac{t^{n+1}}{n+1}+c \\ \Large \int t^3 dt = \dfrac{t^{3+1}}{3+1}+c\)

got it ?

Answer 10

ooo yeah
got it now !!

Answer 11

so : \[6\frac{ t^0 }{ 1 }\] ?

Answer 12

6t^1/1

Answer 13

which is 6t

Answer 14

\(\\ \Large \int 4t^3+6t^0 dt = 4\dfrac{t^{3+1}}{3+1}+6 \dfrac{t^1}{1}c = 4 \dfrac{t^4}{4}+6t+c\\ \Large = t^4+6t+c\)

Answer 15

now substituting the limits (we don't consider the constant 'c')
\(\Large [t^4+6t]^3_{-2} = ...\)

Answer 16

first plug in the upper limit
then lower limit and
subtract

\(\(\Large [t^4+6t]^3_{-2} = .[(3^4+6\times 3) - ((-2)^4+6\times (-2))] =...\)

Answer 17

ignore that \( in the beginning

Answer 18

95 !!!

Answer 19

thank youuuuu !! i appreciate the help!!

Answer 20

welcome ^_^