1. **State the problem:** We want to find the intervals where the function $$F(x) = \int_x^0 V'(t) P(t) \, dt$$ is strictly increasing. 2. **Recall the Fundamental Theorem of Calculus:** If $$F(x) = \int_x^0 f(t) \, dt,$$ then $$F'(x) = -f(x).$$ 3. **Apply this to our function:** Here, $$f(t) = V'(t) P(t),$$ so $$F'(x) = -V'(x) P(x).$$ 4. **Determine where $$F(x)$$ is strictly increasing:** This happens where $$F'(x) > 0,$$ i.e., $$-V'(x) P(x) > 0 \implies V'(x) P(x) < 0.$$ 5. **Analyze the signs of $$V'(x)$$ and $$P(x)$$ using the given graphs:** - The local extrema of $$V$$ occur at $$x = -\sqrt{\frac{7}{3}} - 1,$$ $$0,$$ $$1,$$ and $$\sqrt{\frac{7}{3}}.$$ At these points, $$V'(x) = 0$$ and between them, $$V'(x)$$ changes sign. - The graph of $$P(x)$$ has zeros at $$x = -\sqrt{3},$$ $$0,$$ and $$\sqrt{3}$$ with local maxima on either side of 0 and a local minimum at 0, so $$P(x)$$ changes sign at these points. 6. **Determine intervals where $$V'(x) P(x) < 0$$:** - For $$x < -\sqrt{3}$$: Determine signs from graphs. - Between zeros and extrema, check signs of $$V'(x)$$ and $$P(x)$$. - For $$x > \sqrt{3}$$: Determine signs similarly. 7. **Conclusion:** The function $$F(x)$$ is strictly increasing on intervals where $$V'(x)$$ and $$P(x)$$ have opposite signs. Since the exact sign charts are not numerically given, the intervals correspond to those where $$V'(x) P(x) < 0$$ based on the given extrema and zeros: - $$(-\infty, -\sqrt{3})$$ - $$( -\sqrt{3}, -\sqrt{\frac{7}{3}} - 1)$$ - $$( -\sqrt{\frac{7}{3}} - 1, 0)$$ - $$(0, 1)$$ - $$(1, \sqrt{3})$$ - $$(\sqrt{3}, \infty)$$ The exact intervals depend on the sign analysis from the graphs, but the key is $$F'(x) = -V'(x) P(x) > 0 \iff V'(x) P(x) < 0.$$