Please help! Will FAN AND MEDAL! F of x equals the integral from 1 to x of the natural logarithm of t squared. Use your calculator to find F ′(3). a. 2.197 b. 2.079 c. 1.099 d. 0.693
Can you please help? @agent0smith @jim_thompson5910 @mathmale @Directrix @Kainui @Photon336
@jim_thompson5910 is your go to guy
Use the fundamental theorem of calculus (FTC) If \[\Large f(x) = \int_{a}^{x}g(t)dt\] then \[\Large f \ '(x) = g(x)\]
So this means \[\Large f(x) = \int_{1}^{x}\ln(t^2)dt\] \[\Large f \ '(x) = \ln(x^2)\]
afterwards, just plug in x = 3
I got 2.197? right?
correct
Can you help me with two more?
sure
Thank you. The graph of y = f ′(x) is shown below. List the intervals where the graph of f is concave up. I don't understand what concave up really means?
on a concave up region, the tangent slopes are below the curve on a concave down region, the tangent slopes are above the curve http://mathsfirst.massey.ac.nz/Calculus/Sign2ndDer/images/ConcaveUpDown.png
one way to think of it is like this Feeling down means you have a frowny face so a concave down region looks like a sad frowny face |dw:1463104099741:dw| while feeling up and positive means you have a happy face concave up region has a happy smiley face look |dw:1463104132940:dw|
This is the graph and these are the answers. a. (-4,-2) b. (-2, 2) c. (-4, 0) d. (0, 4)
So a concave up would be c. (-4, 0)? right?
Rule: if `f ' (x)` is increasing on some interval, then `f '' (x) > 0` making `f(x) concave up` on this same interval if `f ' (x)` is decreasing on some interval, then `f '' (x) < 0` making `f(x) concave down` on this same interval
look at the graph of `f ' (x)` on what interval(s) is the function increasing?
(-2,2)
yep `-2 < x < 2`
so f(x) is concave up on the interval -2 < x < 2
so that would be where it concaves up? So the correct answer is b. (-2,2)
yes `-2 < x < 2` in interval notation would be `(-2,2)`
One more question, if you don't mind?
go ahead
A particle moves on the x-axis so that its position is continuous on the interval [3, 13] with some of its values for its velocity v(t) given in the table below. Use a trapezoidal sum with 4 intervals to approximate the total distance the particle travelled in the time interval [3, 13]. |dw:1463104591576:dw| a. 277.50 b. 138.75 c. 115.50 d. 87.50
Step 1) Plot the points on the same coordinate grid. See attached.
Okay. I understand that.
Step 2) Form trapezoids between the points like you see in the attached image.
Step 3) Find the area of each trapezoid. Use this formula \[\Large A = \frac{h*(b_1+b_2)}{2}\] A = area h = height b1 = base1 b2 = base2 b1 and b2 are parallel to each other |dw:1463105174730:dw|
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