Express the area of the region under the curve
y = 6x^3 + 7x^2
and above the x-axis in the form
B
(Integral) f(x) dx
A

Question

Express the area of the region under the curve 
y = 6x^3 + 7x^2
 and above the x-axis in the form 
   B
   (Integral) f(x) dx
   A

anonymous · Answer

I missed this class so I'm lost on how to even approach it :/

anonymous · Answer

This is a nice question, I like it. 

Okay to get where it is above the x-axis, you would need to see where it touches the x-axis. 

Did you graph the function, visualization really helps in this problem.

anonymous · Answer

I'll graph to right now!

anonymous · Answer

Okay the points that touch the x-axis is when y = 0, so here we would let y= 0 and the x-values are actually the values that lies on the axis.

anonymous · Answer

the zero of the function is x=-1.166667

anonymous · Answer

And 0

anonymous · Answer

Yes, that is correct. What you got is right. 

So now if you look at the region under the cure but above the x-axis, it would be from -1.66667 to 0.

So in the case b = -1.66667 and a = 0. 
 and f(x) = function given. 

Hope this helps.

anonymous · Answer

ohhh okay thank you so much!

anonymous · Answer

i see to have gotten it wrong ;/

anonymous · Answer

seem*

anonymous · Answer

$$\int_{-\frac{7}{6}}^0 \left(6 x^3+7 x^2\right)
   \, dx=\frac{2401}{2592}=0.93$$