This is on the 2008 Free response AP Calc AB exam- #4a.
Let f be the function given by f(x)=(lnx)(sinx). The figure above shows the graph of f for 0<=x<=2pi. The function g is defined by int{1,x} f(t)dt for 0<x<=2pi.
a) find g(1). I don't get why it is equal to zero. Like is there a way to tell from the graph or do it analytically?
well, g(1) = int{1,1} f(t)dt
if the integral is done over an interval of zero width, its value is zero.
To find g(1), we need to evaluate the definite integral of f(t) from 1 to 1.
Analytically, we can calculate this integral using the fundamental theorem of calculus. According to the fundamental theorem of calculus, if F(x) is an antiderivative of f(x), then the definite integral of f(x) from a to b is equal to F(b) - F(a).
In this case, we're given that f(x) = (ln(x))(sin(x)). To find an antiderivative, we can use integration by parts.
Using the formula for integration by parts, which states ∫ u dv = uv - ∫ v du, let's assign u = ln(x) and dv = sin(x) dx.
Differentiating u, we get du = (1/x) dx.
Integration of dv, we get v = -cos(x).
Applying integration by parts, we have:
∫ (ln(x))(sin(x)) dx = -ln(x)cos(x) - ∫ (-cos(x))(1/x) dx.
Now, we need to calculate the second part of the integral, which is ∫ (-cos(x))(1/x) dx.
This integral is not elementary, meaning it cannot be expressed in terms of elementary functions. Therefore, we cannot find a closed form for the integral.
Now, let's evaluate the definite integral from 1 to 1:
∫[1,1] (ln(x))(sin(x)) dx = [-ln(x)cos(x) - ∫ (-cos(x))(1/x) dx] evaluated from 1 to 1
When we evaluate the integral from 1 to 1, both limits are the same, so the result is 0.
Therefore, g(1) = 0.
From the graph, it is difficult to determine the exact value of g(1). The graph gives us an idea of the behavior of the function, but to find the exact value, we need to evaluate the integral analytically.
To find g(1), we need to evaluate the integral of f(t) from 1 to 1.
Analytically, the integral of f(t) from 1 to 1 is equal to zero because the limits of integration are the same. If we evaluate the definite integral of a function over the same interval, the result is always zero. So, g(1) = 0.
Looking at the graph, we can also see that the curve representing f(x) (the function lnx*sinx) is symmetric about the y-axis in the interval 0 <= x <= 2pi. This implies that the total area above and below the x-axis between the limits 0 and 2pi should cancel out. Since we are calculating the definite integral over a single point (1 to 1), the area will be zero.
In summary, both analytically and visually, we can conclude that g(1) = 0.