Note: for the following problems... [ b∫a ] should be read as "the intergral over the interval from a to b" or as pictured like this...
1. Find g'(x):
a) g(x) = [x∫0] sin(t) dt
g'(x) = sinx
- We get this answer because we know and committed to memory as to what the antiderivative functions are for trig. functions which we learned are also 'transcendental' functions (don't mind the big word, isn't much useful) and we get -cosx. Then we use the Fundamental Theorum of Calculus Part 1... [ b∫a ] f'(t) = F(b) - F(a) because we're trying to find the definite intergral because when it is over an interval, we're looking for a number. And when there is no interval, then it turns out to be an indefinite integral.
- Using the FTC-part 1, we get g(x) = .
- Simplified g(x) = -cosx + 1.
- Then we simply differentiate g(x) for g'(x)
From this problem, we find out that the derivative of the intergral function, is the same as the underlying function of the intergral. This is true for all functions and is interpreted through the FTC-part 2...
The reason why the variable changes is because, we are only looking and changing x as the other variable t only changes as x does.
b) g(x) = [x∫0] √(1+t3) dt
g'(x) = √(1+x3)
- Using the FTC-part2. (See the pattern?)
c) g(x) = [3x²∫0] sin(t) dt
g'(x) = sin(3x2) · 6x
- The difference about this question to the first is the variable x has changed.
- We interpret this as a composite of functions, and then use the chain rule as we normally do to differentiate these functions...
g'(x) = f'(g(x)) · g'(x)
d) g(x) = [sinx∫0] √(1+t3) dt
g'(x) = √(1+sin3x) · -cosx
- Using FTC-part 2 and the chain rule law.
2. Graph of f
a) Find g(1), g'(1), g''(1)
g(1) = 1.5
g'(1) = 0
g''(1) = -3
- g(1) is found by finding the sign area under the curve from 0 to 1, (f(1)·1)/2 'because it's a triangle.
- g'(1) is found by reading straight off from the graph, as the FTC-part 2 proves that a derivative of an intergrated function is just its underlying function.
- g''(1) is found by finding the slope of the tangent line at g'(1)
b) For what values of x on the interval (-2,2) is g increasing? Explain your reasoning.
Since g'(x) = f(t), we just look at where ever it is positively valued on the graph.
c) For what values of x, on the interval (-2,2) is g concave down?
Since g'(x) = f(t), we just look at where ever its slope is negatively valued on the graph.
d) Sketch the graph of g
This is just one of the possible graphs for g as it is interpreted as a parent function derived from its derivative.
The graph of g'(x)=f(t), and the derivative only tells you the shape of the parent graph not where it is vertically. "If you were given a point on the graph, you would be able to position it, but if not, there would be infinitely many answers. (e's correction)" I got this answer by looking at the derivative and second derivative for increasing/decreasing and concavity characteristics. Aswell as integrating the function from -2 to 2, to find the total change for more accuracy. Using different intervals when intergrating will give you the same shape of the graph but it will be positioned different vertically.
Take care everyone!
Next scribe will be