#$&*
course Mth 174
10/10/11 406pm
Section 6.41) 6.4.6 Assume that F'(t) = sin t * cos t and F(0) =1. Find F(b) for b = 0, 1/2, 1, 3/2, 2, 5/2, 3.
A: Using u substitution, I found that the anti derivative is ½(sin(x))^2.
@& You still need to find F(0), F(1/2), etc..
Your general antiderivative includes an integration constant c.
The condition F(0) = 1 dictates the value of that constant. Once you've evaluated the constant, you can write the particular function that satisfies these conditions, and evaluate it at 0, 1/2, etc..
Alternatively, if you evaluate 1/2 (sin(x))^2 at two values of x, you find how much F(t) changed between these values. You can use this to evaluate the function, given that F(0) = 1.*@
2) 6.4.8 Write an expression for the function f(x) with the given properties. f'(x) = (sin x)/x and f(1) = 5.
A: The fundamental theorem of calculus states that F(a)-F(b)=int f(x)dx so f(x)= 5+int(sinx)/xdx
@& F(a) - F(b) is the value of the definite integral int( sin(x) / x dx, x from a to b).*@
3) 6.4.12 Find d/dx(Int(ln(t) dt, x, 1)).
A: Because a=x and b=1, the answer is -ln(x). If a=1 and b=x, then the answer would have been ln(x).
Explanation of notation: int( f(x) dx, a, b) means the integral of f(x) with respect to x, from x = a to x = b.
4) 6.4.18 Find d/dx(Int(e^-(t^2) dt, 0 , x^3)).
A: Using u substitution for u=x^3, and substituting that u for t we get 3x^2*e^(x^6).
5) 6.4.22 Let F(x) = Int(sin(4t) dt, 0, x). Evaluate F(pi). Draw a sketch and describe geometrically why your value for F(pi) is correct. For what values of x is F(x) positive and for which value is it negative?
A: The anti derivative of sin(4t) is -cos(4t)/4 so between 0 and pi we can say that F(pi) is -cos(4pi)/4+cos(0)/4=-1/4+1/4.=0
F(x) is positive for x<=pi/4
negative for x>=pi/4
The graph is curve that is a wave on the x axis and F(pi) is 0 because the area under the x axis is equal to the area above the x axis before x=pi.
@& Good, but check my notes on #1.*@