1. **Problem statement:** Find the area of the region enclosed by the graphs of $f(x) = \cos(x^2)$ and $g(x) = e^x$ for $-1.5 \leq x \leq 0.5$. 2. **Formula and approach:** The area between two curves $f(x)$ and $g(x)$ from $a$ to $b$ is given by $$\text{Area} = \int_a^b |f(x) - g(x)| \, dx$$ We need to determine which function is on top in the interval and integrate the difference accordingly. 3. **Determine which function is greater:** Evaluate at a test point, say $x=0$: $$f(0) = \cos(0^2) = \cos(0) = 1$$ $$g(0) = e^0 = 1$$ At $x=0$, both are equal. Check at $x=-1$: $$f(-1) = \cos(1) \approx 0.5403$$ $$g(-1) = e^{-1} \approx 0.3679$$ So $f(x) > g(x)$ at $x=-1$. Check at $x=0.5$: $$f(0.5) = \cos(0.25) \approx 0.9689$$ $$g(0.5) = e^{0.5} \approx 1.6487$$ So $g(x) > f(x)$ at $x=0.5$. 4. **Find intersection points:** Solve $\cos(x^2) = e^x$ in $[-1.5,0.5]$. By inspection, $x=0$ is an intersection. 5. **Set up the integral:** From $-1.5$ to $0$, $f(x) \geq g(x)$, so area is $$\int_{-1.5}^0 (\cos(x^2) - e^x) \, dx$$ From $0$ to $0.5$, $g(x) \geq f(x)$, so area is $$\int_0^{0.5} (e^x - \cos(x^2)) \, dx$$ 6. **Final area:** $$\text{Area} = \int_{-1.5}^0 (\cos(x^2) - e^x) \, dx + \int_0^{0.5} (e^x - \cos(x^2)) \, dx$$ This integral does not have a simple closed form and is best evaluated numerically. --- **Slug:** area enclosed **Subject:** calculus **Desmos:** {"latex":"y=\cos(x^2), y=e^x","features":{"intercepts":true,"extrema":true}} **q_count:** 2