- #1
courtrigrad
- 1,236
- 2
Hello all
Here are a few problems I encountered:
Find the area bounded by the parabola [tex] y = \frac{1}{2}x^2 + 1 [/tex] and the straight line [tex] y = 3 + x [/tex] Ok, so I first set the two equations equal to each other:
[tex] 3 + x = \frac{1}{2}x^2 + 1 [/tex] After solving for x, do I just use those two values for the upper and lower limits, and evaluate:
[tex] \int^b_a \frac{1}{2}x^2 + x - 2 [/tex]?
Also how would you evalutate the following integrals:
[tex] \int^b_a (x+1)^a dx [/tex]
[tex] \int^b_a sin \alpha x {} dx [/tex]
[tex] \int^b_a cos \alpha x {} dx [/tex]
For the first one would I use a geometric progression? I know that for [tex] \int^b_a x^a dx [/tex] you divide up the interval using the following points:
[tex] a, aq, aq^2, . . . , aq^n^-1, aq^n = b [/tex] So that means for [tex] a+1 [/tex] we have [tex] a+1, (a+1)q, (a+1)q^2, . . . (a+1)q^n-1, (a+1)q^n [/tex]?
For the other integrals, do I just make use of the identity [tex] 2\sin u sin v = \cos(u-v) - \cos(u+v) [/tex]?
Thanks
Here are a few problems I encountered:
Find the area bounded by the parabola [tex] y = \frac{1}{2}x^2 + 1 [/tex] and the straight line [tex] y = 3 + x [/tex] Ok, so I first set the two equations equal to each other:
[tex] 3 + x = \frac{1}{2}x^2 + 1 [/tex] After solving for x, do I just use those two values for the upper and lower limits, and evaluate:
[tex] \int^b_a \frac{1}{2}x^2 + x - 2 [/tex]?
Also how would you evalutate the following integrals:
[tex] \int^b_a (x+1)^a dx [/tex]
[tex] \int^b_a sin \alpha x {} dx [/tex]
[tex] \int^b_a cos \alpha x {} dx [/tex]
For the first one would I use a geometric progression? I know that for [tex] \int^b_a x^a dx [/tex] you divide up the interval using the following points:
[tex] a, aq, aq^2, . . . , aq^n^-1, aq^n = b [/tex] So that means for [tex] a+1 [/tex] we have [tex] a+1, (a+1)q, (a+1)q^2, . . . (a+1)q^n-1, (a+1)q^n [/tex]?
For the other integrals, do I just make use of the identity [tex] 2\sin u sin v = \cos(u-v) - \cos(u+v) [/tex]?
Thanks
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