How to solve with integration?

how to do question 10 a) b) and question 3 and 4?

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3 Answers
May 13, 2018

#Q=(15/2,0)#
#P=(3,9)#
#"Area"=117/4#

Explanation:

Q is the x-intercept of the line #2x+y=15#

To find this point, let #y=0#
#2x=15#
#x=15/2#

So #Q=(15/2,0)#

P is a point of interception between the curve and the line.

#y=x^2" "(1)#
#2x+y=15" "(2)#

Sub #(1)# into #(2)#

#2x+x^2=15#
#x^2+2x-15=0#
#(x+5)(x-3)=0#
#x=-5# or #x=3#

From the graph, the x co-ordinate of P is positive, so we can reject #x=-5#

#x=3#
#y=x^2#
#=3^2#
#=9#

#:. P=(3,9)#

graph{(2x+y-15)(x^2-y)=0 [-17.06, 18.99, -1.69, 16.33]}

Now for the area

To find the total area of this region, we can find two areas and add them together.
These will be the area under #y=x^2# from 0 to 3, and the area under the line from 3 to 15/2.

#"Area under curve"=int_0^3 x^2dx#
#=[1/3x^3]_0^3#
#=1/3xx3^3-0#
#=9#

We can work out the area of the line through integration, but its easier to treat it like a triangle.

#"Area under line"=1/2xx9xx(15/2-3)#
#=1/2xx9xx9/2#
#=81/4#

#:."total area of shaded region"=81/4+9#
#=117/4#

May 13, 2018

For 3 & 4

[Tom's done 10]

Explanation:

3

#int_0^5 f(x) \ dx = (int_0^1 + int_1^5) f(x) \ dx#

#:. int_1^5 f(x) \ dx = (int_0^5 - int_0^1) f(x) \ dx #

#= 1- (-2) = 3#

4

#int_(-2)^3 f(x) dx = (int_(-2)^1 + int_1^3) f(x) dx#

#:. int_(3)^(-2) f(x) dx = -int_(-2)^3 f(x) dx #

# = - (int_(-2)^1 + int_1^3) f(x) dx #

#= - (2 - 6) = 4#

May 13, 2018

See below:

Warning: Long answer!

Explanation:

For (3):

Using the property:

#int_a^b f(x) dx=int_a^c f(x) dx + int_c^b f(x) dx#

Hence:

#int_0^5 f(x) dx=int_0^1 f(x) dx + int_1^5 f(x) dx#

#1=-2 + x#

#x=3=int_1^5 f(x) dx#

For (4):

(same thing)

#int_a^b f(x) dx=int_a^c f(x) dx + int_c^b f(x) dx#

#int_-2^3 f(x) dx=int_-2^1 f(x) dx + int_1^3 f(x) dx#

#x=2+(-6)#

#x=-4=int_-2^3 f(x) dx#

However, we must swap the limits on the integral, so:
#int_3^-2 f(x) dx=-int_-2^3 f(x) dx#

So:#int_3^-2 f(x) dx=-(-4)=4#

For 10 (a):

We have two functions intersecting at #P#, so at #P#:

#x^2=-2x+15#

(I turned the line function into slope-intercept form)

#x^2 + 2x-15=0#

#(x+5)(x-3)=0#

So #x=3# as we to the right of the #y# axis, so #x>0#.

(inputting #x=3# into any of the functions)

#y=-2x+15#

#y=-2(3)+15#

#y=15-6=9#

So the coordinate of #P# is #(3,9)#

For #Q#, the line #y=-2x+15# cuts the #y#-axis, so #y=0#

#0=-2x+15#

#2x=15#

#x=(15/2)=7.5#

So #Q# is located at #(7.5, 0)#

For 10 (b).

I will construct two integrals to find the area. I will solve the integrals separately.

The area is:
#int_a^b f(x) dx=int_a^c f(x) dx + int_c^b f(x) dx#

#A=int_O^Q f(x) dx=int_O^P (x^2) dx + int_P^Q (-2x+15) dx#

(Solve first integral)

#int_O^P (x^2) dx=int_0^3 (x^2) dx=[x^3/3]#

(substitute the limits into the integrated expression, remember:
Upper-lower limit to find the value of integral)

#[3^3/3]-0=9=int_O^P (x^2) dx#

(solve second integral)

#int_P^Q (-2x+15) dx=int_3^7.5 (-2x+15) dx=[(-2x^2)/2+15x]=[-x^2+15x]#

(substitute limits: Upper-lower)

#[-(15/2)^2+15(15/2)]-[-3^2+15(3)]#

#[(-225/4)+(225/2)]+[9-45]=[(-225/4)+(450/4)]+[-36]= [(225/4)]+[(-144/4)]=(81/4)#

#int_P^Q (-2x+15)dx=(81/4)#

#int_O^Q f(x) dx=int_O^P (x^2) dx + int_P^Q (-2x+15) dx#

#A=int_O^Q f(x) dx=9 + (81/4)#

#A=int_O^Q f(x) dx=9 + (81/4)#

#A=(36/4)+(81/4)#

#A=(117/4)#