I'm trying to make use of the definite integral property
My equation was x^2dx evaluated between 2 and -2.
Here is my math. Let me know if it's right.
x^2dx=1/3 x^3
evaluated: (1/3 2^3)-(1/3 (-2)^3)=8/3+8/3=16/3
Did i do this correctly?
2007-07-24 08:02:55 · 9 answers · asked by t b 1 in Science & Mathematics ➔ Mathematics
I = ∫ x ² dx between given limits
I = x ³ / 3 between limits
I = 2 ³ / 3 - (- 2 ³ / 3)
I = 8 / 3 + 8 / 3
I = 16 / 3 (which agrees with your answer)
2007-07-24 21:46:46 · answer #1 · answered by Como 7 · 0⤊ 0⤋
you have to be careful about the limits of integration
if its -2 to 2 (the positive 2 is on the top) then you did it right and the answer is 16/3
however if its 2 to -2 like you typed ( the -2 is on the top) then the answer is -16/3
2007-07-24 08:11:09 · answer #2 · answered by joecool92586 1 · 0⤊ 0⤋
Yes, but you might want to investigate alternate ways of evaluating definite integrals such as this one based on the symmetry of the functions graph. Because the graph is symmetrical to the y-axis, you can evaluate by:
2 * def. integral 0,2 as the areas from -2,0 and 0,2 are equal.
2007-07-24 08:14:02 · answer #3 · answered by gfulton57 4 · 0⤊ 0⤋
I worked out the problem and my work looks just like yours, with the answer 16/3. You seem to have a firm grasp of this property, good job.
2007-07-24 08:11:46 · answer #4 · answered by EM 2 · 0⤊ 0⤋
Correct. Just worked it myself, and I got exactly the same thing as you! Good work.
2007-07-24 08:09:29 · answer #5 · answered by victoria 5 · 0⤊ 0⤋
Very nice work. That is exactly correct.
2007-07-24 08:09:27 · answer #6 · answered by C-Wryte 4 · 0⤊ 0⤋
Yes, good job!
2007-07-24 08:09:02 · answer #7 · answered by whitesox09 7 · 0⤊ 0⤋
That is right.
2007-07-24 08:44:34 · answer #8 · answered by beaufordrenee 2 · 0⤊ 0⤋
its correct
2007-07-24 08:13:49 · answer #9 · answered by Friend 3 · 0⤊ 0⤋