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Calculus1 8 Online
OpenStudy (anonymous):

plz ,may anyone tell me what does it mean when u integrate a function to find the value of an area and u get a negative value?

OpenStudy (anonymous):

If the integral over some interval is negative, it means there is more area below the x axis than above. Remember that when calculating integrals, area below the x axis is negative.

OpenStudy (anonymous):

for this case y= f(x)was given as-2x^2-5x+6 and the x axis -4<x<0!i am not sure as well that i followed the right steps.

OpenStudy (anonymous):

So are you trying to solve this integral? \[\int\limits_{-4}^{0} (-2x^2 -5x +6) dx\]

OpenStudy (anonymous):

yep?in fact i like this thing but i am not good at it!

OpenStudy (anonymous):

Ok. To find the antiderivative of this, we use one of the most basic integration rules, which states the following: \[\int\limits_{?}^{?} x^{a} = \frac{ 1 }{ a+1 } x^{a+1}\]Heres what this is saying. To integrate, we add 1 to the exponent, then multiply that by the reciprocal of the new exponent. I'll integrate the first term for you to get you started: \[\int\limits_{?}^{?} -2x^2 = \frac{ 1 }{ 2+1 } -2 x^{2+1}=\frac{ -2 }{ 3 }x^3\]

OpenStudy (anonymous):

i got it!very help ful ,but a bit different from simpler methods where i got confused from!

OpenStudy (anonymous):

The power rule is absolutely the method you'll use to integrate this. You're basically breaking the integral into three parts, one term per part. Integrate each part to get your antiderivative. Once you get that function, plug in 0, plug in -4, and subtract the second from the first.

OpenStudy (anonymous):

all right sir!i got 86/3units^2,and i think thats correct!what about y=(x+2)^2 and x+y=7area at 0<x<5?

OpenStudy (anonymous):

If you're looking for the area between two graphs, you need to figure out where they intersect to figure out the interval we need to use. Then you need to figure out which graph is above and which is below. Then, subtract the lower graph from the upper graph to get your new function. Then integrate as usual.

OpenStudy (anonymous):

plz how to find out where they intercept

OpenStudy (anonymous):

Set the two functions equal to each other and solve for x, or use a graphing calculator

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