I Can't Intergrate It? (1 Viewer)

[BigBear_25]

Hi ppl.

I got this really annoying intergration question.

Find the area enclosed by the curve y=x^3 and
the line y=-3x+4.

Can u plz help

Thanks.

PS: I hate intergration.

 

[Riviet]

It's "integration" lol.

Find the area enclosed by the curve y=x^3 and
the line y=-3x+4

You will need to find the point of intersection of the two curves by solving y=x³+3x-4, find the area between y=x³ and x-axis from 0 to x=(insert point of intersection here) and add this to area between y=-3x+4 and x-axis from x=(insert point of intersection here) to x=4.

Edit: the point of intersection turned out to be 1 LOL!

(Cubic Equation Calculator Here)

 

[Mountain.Dew]

uhhhh i dont think u can find the area enclosed, or bound by these two curves. there is only one pt of intersection, not two.

consider x^3 +3x - 4 = 0, solve for x (in real number system)

x = 1 works

so we get (by long division) x^3 + 3x - 4 = (x-1)(x^2 + x + 4)

x^2 + x + 4 has no real roots (discriminant < 0 )

there arent two pts of intersection to enclose an area with these curves

hence, the only correct answer is undefined, or, infinity.

however, if it said bound by the curves and a certain axis, then it is doable.