1. **State the problem:** We are given the function $$f(t) = 4 - 4t^2 + 4t^4 - 4t^6 + \cdots = \sum_{n=0}^\infty (-1)^n 4 t^{2n}$$ and the function $$F(x) = \int_0^x f(t) \, dt.$$ We want to find an expression for $$F(x)$$ and understand its behavior. 2. **Recognize the series:** The series for $$f(t)$$ is a power series with general term $$4(-1)^n t^{2n}$$. This can be factored as $$4 \sum_{n=0}^\infty (-1)^n (t^2)^n$$. 3. **Use the geometric series formula:** Recall that for $$|r| < 1$$, $$\sum_{n=0}^\infty r^n = \frac{1}{1-r}$$. Here, $$r = -t^2$$, so for $$|t| < 1$$, $$f(t) = 4 \sum_{n=0}^\infty (-t^2)^n = \frac{4}{1 + t^2}.$$ 4. **Rewrite $$F(x)$$:** Using this closed form for $$f(t)$$, $$F(x) = \int_0^x \frac{4}{1+t^2} \, dt.$$ 5. **Integrate:** The integral of $$\frac{1}{1+t^2}$$ is $$\arctan(t)$$, so $$F(x) = 4 \int_0^x \frac{1}{1+t^2} \, dt = 4 [\arctan(t)]_0^x = 4 \arctan(x).$$ 6. **Summary:** - The original series converges to $$\frac{4}{1+t^2}$$ for $$|t|<1$$. - The integral $$F(x)$$ is $$4 \arctan(x)$$. This means $$F(x)$$ is a smooth function increasing from 0 at $$x=0$$ to $$4 \times \frac{\pi}{2} = 2\pi$$ as $$x \to \infty$$.