Could you please answer this question below?

enter image source here

2 Answers
Dec 3, 2017

B. #-1.2#

Explanation:

The integral #int_-2^5f(x)dx# represents the area under the curve. Since we are given the areas of the regions we can calculate the area but must distinguish the positive area and negative area.

enter image source here

Basically the area above the #x# axis is positive while the area below the #x# axis is negative.

Looking back that the problem lets mark the positive and negative areas.

enter image source here

So we calculate the area:

#A=3.6-3.0-1.8=-1.2#

Dec 3, 2017

#A=17.15#

None of the answers given in #A# through #E# is correct. Neither are the values of each segment of the area as shown. I have checked it against two online integral calculators and both gave me the answer I got.

Explanation:

.
From the graph we can see that the roots of #f(x)# are:

#x=-2, 0,3,3,and 5#

The reason for listing #3# twice is that the graph touches the #x#-axis but does not cross it (multiplicity of roots).

As such, we can write the function for #f(x)#:

#f(x)=(x+2)(x)(x-3)(x-3)(x-5)#

#f(x)=(x^2+2x)(x^2-6x+9)(x-5)#

#f(x)=(x^4-6x^3+9x^2+2x^3-12x^2+18x)(x-5)#

#f(x)=(x^4-4x^3-3x^2+18x)(x-5)#

#f(x)=x^5-5x^4-4x^4+20x^3-3x^3+15x^2+18x^2-90x#

#f(x)=x^5-9x^4+17x^3+33x^2-90x#

Now, to find the area between the curve and the #x#-axis between #x=-2 and 5#, we will take the integral of the functions and evaluate it between those two limits:

#A=int_-2^5(x^5-9x^4+17x^3+33x^2-90x)dx#

#A=1/6x^6-9/5x^5+17/4x^4+11x^3-45x^2# evaluated between limits of #-2 and 5#

#A=(1/6*5^6-9/5*5^5+17/4*5^4+11*5^3-45*5^2)-(1/6*(-2)^6-9/5*(-2)^5+17/4*(-2)^4+11*(-2)^3-45*(-2)^2)#

#A=(2604.17-5625+2656.25+1375-1125)-(10.67+57.6+68-88-180)#

#A=-114.58-(-131.73)=-114.58+131.73=17.15#