 #1
chwala
Gold Member
 2,210
 286
 Homework Statement:

find the definite integral of
##\int_0^5 x^3+3x^2+6x8\,dx##
 Relevant Equations:
 integration
##\int_0^5 [x^3+3x^2+6x8\,]dx##
##\int_0^1 [x^3+3x^2+6x8\,]dx= \frac {17}{4}##
##\int_1^4 [ x^3+3x^2+6x8\,]dx= 16##
##\int_4^5[x^3+3x^2+6x8\,]dx= \frac {49}{4}##
Therefore, total area is ##\frac {17}{4}+ 16+\frac {49}{4}=32.5##
now where my problem is,... my colleague indicated the answer as ##36.75##
who is right, or where have i made a blunder?
##\int_0^1 [x^3+3x^2+6x8\,]dx= \frac {17}{4}##
##\int_1^4 [ x^3+3x^2+6x8\,]dx= 16##
##\int_4^5[x^3+3x^2+6x8\,]dx= \frac {49}{4}##
Therefore, total area is ##\frac {17}{4}+ 16+\frac {49}{4}=32.5##
now where my problem is,... my colleague indicated the answer as ##36.75##
who is right, or where have i made a blunder?
Last edited: