area under the curve

[bcddd214]

need a check. I hope I am at least close.

Determine the area under the curve of f(x)=-x^2+2x+3 for-1?x?3
?_(-1)^3??-x+2x+3 dx/dt?
B=1/3 (3) 3^3+1/2 (4) 3^3+(3)3=27+9+9=45
A=1/3 (3) ?-1?^3+1/2 (4) ?-1?^3+(3)-1=.99-2+2=.99

 

[tkhunny]

Unable to translate.

 

[bcddd214]

Unable to translate.

find a book then. I read this loud and clear.

 

[tkhunny]

Learning to communicate is part of the mathematics experience.

 

[galactus]

That is difficult to read. If you're going to be a regular on the site, perhaps try learning a little LaTex.

Find area under the curve \(\displaystyle f(x)=-x^{2}+2x+3, \;\ -1\leq x\leq 3\)

\(\displaystyle \int_{-1}^{3}(-x^{2}+2x+3)dx=\left \frac{-1}{3}x^{3}+x^{2}+3x\right|_{-1}^{3}=\frac{32}{3}\)

Click on 'quote' at the upper right corner of this post to see what I typed to make it display this way.

 

[lookagain]

I reported user bcddd214 to both tkhunny and to Ted.