blogggggggggyyyyyyyyyy again!
1. The graph of y=5x^4-x^5 has an inflection point or points at
y=5x^4-x^5
20x^3-5x^3
20x^2(3-x)
3-x=0
-x=-3
divide by -1
x=3
2. d/dx(integral form 0 to x^2)sin^2t(dt)
sin^2(x^2)
2xsin^2(x^2)
3. The average value of f(x)=1/x from x=1 to x=e is
(1/e-1)[ln e -ln 1] *the e and 1 are the absoule value of
(1/e-1)[1-0]
= 1/e-1
4. Find the area under the curve y=2x-x^2 from x=1 to x=2 with n=4 left-endpoint rectangles.
use LRAM
delta x=2-1/4=1/4
1/4[f(1)+(5/4)+f(6/4)+f(7/4)]
1/4(0+.4375+.75+.9375)
1/4(2.125)= 17/32
i need help on SUBSTITUION, i keep forgetting ahhh
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When using substitution, you should find your u and du first. Once you have done that, you manipulate the problem to satisfy du (usually you have to put a fraction in the front) and then plug in u. After that, you integrate u, plug back in for u and then solve. Hope this helps!
ReplyDeleteSubstitution steps:
ReplyDelete1. Find the derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
Substitution: usually let u = a function that’s “inside” another function, especially if du (possibly off
ReplyDeleteby a multiplying constant) is also present in the integrand.