Question

Find the accumulation function \(F\). Then evaluate \(F\) at each value of the independent variable and graphically show the area given by each value of \(F\). $$ F(\alpha)=\int_{-1}^{\alpha} \cos \frac{\pi \theta}{2} d \theta \quad \text { (a) } F(-1) \quad \text { (b) } F(0) \quad \text { (c) } F\left(\frac{1}{2}\right) $$

Short answer

\(F(-1) = 0\), \(F(0) = \frac{2}{\pi}\) and \(F(\frac{1}{2}) = \frac{2}{\pi}(1+ \sqrt{2}/2)\)

Step-by-step solution

Compute the accumulation function F(α)

Based on the fundamental theorem of calculus, we compute the integral from -1 to α of \(\cos(\frac{π\theta}{2})\) dΘ. When we do this, we get: \(F(α) = \frac{2}{π}[\sin(\frac{πα}{2}) + \sin(\frac{π}{2})] \)

Evaluate F(α) at α=-1

Substitute α = -1 into \(F(α)\). So we have \(F(-1) = \frac{2}{π}[\sin(-\frac{π}{2}) + \sin(\frac{π}{2})] = 0\)

Evaluate F(α) at α=0

Substitute α = 0 into \(F(α)\). This gives: \(F(0) = \frac{2}{π}[\sin(0) + \sin(\frac{π}{2})] = \frac{2}{π}\)

Evaluate F(α) at α=1/2

We then substitute α = 1/2 into \(F(α)\). This gives: \(F(\frac{1}{2}) = \frac{2}{π}[\sin(\frac{π}{2}) + \sin(\frac{π}{4})] = \frac{2}{π}(1+ \sqrt{2}/2)\)

Graphical Representation

Plot these points (-1, 0), (0, 2/π), (1/2, 2/π * (1+sqrt(2)/2)) on a graph. Join them using smooth curves. The graph is the graphical representation of F(α)