The area of the region between the graph of y = 8x^3 + 5x and the x-axis from x = 1 to x = 3 is?

Question

A. 100
B. 140
C. 160
D. 180


Please show me how to get to the answer, and not just the answer. Thank you.

Answer 1

Hi,

The answer is D, 180

To find that, take the integral of 8x^3 + 5x.

It equals 2x^4 + 5/2x^2. Then find the value when x = 3 minus the value when x = 1.

When x = 3:
2x^4 + 5/2x^2 = 2(3)^4 + 5/2*(3)^2 = 2*81 + 5/2 *9 = 162 +22.5 = 184.5

When x = 1:
2x^4 + 5/2x^2 = 2(1)^4 + 5/2*(1)^2 = 2 + 5/2 = 4.5

Then 184.5 - 4.5 = 180. that's your answer.


You can also get this on your TI, if you enter Y1 = 8x^3 + 5x and go to CALC, integral of f(x) dx. Select 1 as your lower limit and 3 as your upper limit and the calculator will also give you 180 as your answer.


I hope that helps!! :-)

Answer 2

You are given 2 equations, y = 8x^3 + 5x and the x- axis, which is y = 0. The first curve is the upper curve and y = 0 is the lower curve. You can test this out by simply plugging in a number to each, in which can you will always get a positive number for the first equation and you will get 0 for the second.

Now you integrate 8x^3 + 5x - 0 from x = 1 to x = 3:

You get
2x^4 + 5/2 x^2 evaluated from 1 to 3:
162 + 45/2 -2 - 5/2 = 184.5 - 4.5 = 180

The answer is D.

Answer 3

A = ∫ 8x³ + 5x dx between lims. 1 to 3
A = [ 2x^(4) + 5x²/2 ] between 1 and 3
A = (2 x 81 + 45/2) - (2 + 5/2)
A = 160 + 20
A = 180
ANSWER D