- #1
Briggs
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I am having a little trouble on this problem for calculus..
Curve has equation y = x³ - 6x² + 9x + 16
I am asked to find [tex]\int_{-1}^{2} (x^3 - 6x^2 + 9x + 16) dx[/tex]
I get the answer to be -3/4 but the book gives the answer as 47.25. I do not know how the book got this. The way I found it was
[tex][\frac{x^4}{4} - \frac{6x^3}{3} + \frac{9x^2}{2}]_{-1}^{2}[/tex] and then putting x=2 and solving it then putting x=-1 and solving it then subtract the answer from x=-1 from x=2 so i found [tex](4-16+18)-(\frac{1}{4}--2+\frac{9}{2}) = \frac{-3}{4}[/tex]
It is important that I get this answer correct because It is needed for the next question of finding the area under a curve, could it just be that the book has an incorrect answer as it has known to be in the past or am i going about this completely the wrong way.
Thanks for any help you guys can provide
Curve has equation y = x³ - 6x² + 9x + 16
I am asked to find [tex]\int_{-1}^{2} (x^3 - 6x^2 + 9x + 16) dx[/tex]
I get the answer to be -3/4 but the book gives the answer as 47.25. I do not know how the book got this. The way I found it was
[tex][\frac{x^4}{4} - \frac{6x^3}{3} + \frac{9x^2}{2}]_{-1}^{2}[/tex] and then putting x=2 and solving it then putting x=-1 and solving it then subtract the answer from x=-1 from x=2 so i found [tex](4-16+18)-(\frac{1}{4}--2+\frac{9}{2}) = \frac{-3}{4}[/tex]
It is important that I get this answer correct because It is needed for the next question of finding the area under a curve, could it just be that the book has an incorrect answer as it has known to be in the past or am i going about this completely the wrong way.
Thanks for any help you guys can provide