How to answer these using intergration ?

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Answer 1 · 2018-03-08T08:39:51

The area is #=(32/3)u^2# and the volume is #=(512/15pi)u^3#

Start by finding the intercept with the x-axis

#y=4x-x^2=x(4-x)=0#

Therefore,

#x=0# and #x=4#

The area is

#dA=ydx#

#A=int_0^4(4x-x^2)dx#

#=[2x^2-1/3x^3]_0^4#

#=32-64/3-0#

#=32/3u^2#

The volume is

#dV=piy^2dx#

#V=piint_0^4(4x-x^2)^2dx#

#=piint_0^4(16x^2-8x^3+x^4)dx#

#=pi[16/3x^3-2x^4+1/5x^5]_0^4#

#=pi(1024/3-512+1024/5-0)#

#=pi(5120/15-7680/15+3072/15)#

#=pi(512/15)#

Answer 2 · 2018-03-08T08:43:25

a. #32/3#

b. #(512pi)/15#

First, we need to find the points at which the graph crosses the #x#-axis.

#4x-x^2=x(4-x)=0#

Either #x=0# or #4-x=0#

#x=0 or 4#

Now we know our upper and lower bounds.

a. #"Area under a graph"=int_b^af(x)dx#

#int_0^4 4x-x^2dx=[2x^2-x^3/3]_0^4=(2(4)^2-4^3/3)-(2(0)^2-0^3/3)=32/3#

b. #"Volume of rotation"=piint_b^a(f(x))^2dx#

#f(x)^2=(4x-x^2)^2=16x^2-8x^3+x^4#

#piint_0^4 16x^2-8x^3+x^4dx=pi[(16x^3)/3-2x^4+x^5/5]_0^4=pi[((16(4)^3)/3-2(4)^4+4^5/5)-((16(0)^3)/3-2(0)^4+0^5/5)]=pi[512/15]=(512pi)/15#