At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Ur attachment has the graph drawn in it...If the graph is correct then its area should give you the integral
how do I find the area????!?!
what the heck how?
Did you factor the conjugal value of the exponent?
it's going from 1 to 5 should it be get the antiderivative of that equation and then f(5)-F(1) ... x.x.x.
??? I don't understand @shaagasda
Usuki, he didn't follow the derivative law for conjugal exponents...
Don't confuse him lol....
The answer is 10.716524....
Pay more attention in class lol
I did plug in all the values of x from 1 to 5 and I did draw the graph. Now, the integral part x.x.
Lol... that's like doing it in long division format... Never mind, I'm done lol.
My answer was correct.
you have the maximum value of the function as 3...and clearly...the eqn is one degree so its the graph of a straight line...Now the function covers area from 1 to 5 only..so shouldn't the area be 1/2*3*4 thats 6??
Try it on your TI-84...
sorry 0.5*3*5=7.5... would that be wrong??..@shaagasda
Fine... I'll take it...
it's called I'm trying to refresh my memory on an old topic!
well then...check the answer and see if it matches
but why is it 0.5X3X5? Am I taking the area of a triangle of something....ooohhh 1/2 (base)(height)