1. **State the problem:** Find the area enclosed between the curves $y = e^x$ and $y = x^2 + 1$. 2. **Formula and approach:** The area between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x=b$ is given by: $$\text{Area} = \int_a^b |f(x) - g(x)| \, dx$$ We first find the points of intersection to determine the limits $a$ and $b$. 3. **Find points of intersection:** Solve $e^x = x^2 + 1$. Check $x=0$: $e^0=1$, $0^2+1=1$ so $x=0$ is an intersection. Check $x=1$: $e^1 \approx 2.718$, $1^2+1=2$, so $e^x > x^2+1$ at $x=1$. Check $x=-1$: $e^{-1} \approx 0.368$, $(-1)^2+1=2$, so $e^x < x^2+1$ at $x=-1$. Check $x=-0.7$: $e^{-0.7} \approx 0.496$, $(-0.7)^2+1=1.49$, so $e^x < x^2+1$. Check $x=0.5$: $e^{0.5} \approx 1.65$, $0.5^2+1=1.25$, so $e^x > x^2+1$. By testing values, the curves intersect at $x=0$ and another point near $x=1.7$ (approximate). Use numerical methods or graphing to find the second intersection $x \approx 1.84$. 4. **Set limits:** $a=0$, $b \approx 1.84$. 5. **Determine which function is on top:** For $x$ in $[0,1.84]$, $e^x > x^2 + 1$. 6. **Set up integral:** $$\text{Area} = \int_0^{1.84} (e^x - (x^2 + 1)) \, dx$$ 7. **Compute integral:** $$\int_0^{1.84} e^x \, dx = e^{1.84} - e^0 = e^{1.84} - 1$$ $$\int_0^{1.84} x^2 \, dx = \frac{(1.84)^3}{3}$$ $$\int_0^{1.84} 1 \, dx = 1.84$$ 8. **Combine results:** $$\text{Area} = (e^{1.84} - 1) - \frac{(1.84)^3}{3} - 1.84$$ 9. **Calculate numerical values:** $$e^{1.84} \approx 6.30$$ $$(1.84)^3 = 1.84 \times 1.84 \times 1.84 \approx 6.23$$ $$\frac{6.23}{3} \approx 2.08$$ 10. **Final area:** $$\text{Area} \approx (6.30 - 1) - 2.08 - 1.84 = 5.30 - 2.08 - 1.84 = 1.38$$ **Answer:** The area enclosed between $y = e^x$ and $y = x^2 + 1$ is approximately $1.38$ square units.