1. Problem 1: Find the area enclosed between the curves $y = x^2$ and $y = 4x - x^2$. 2. First, find the points of intersection by setting $x^2 = 4x - x^2$. 3. Solve for $x$: $$x^2 = 4x - x^2 \implies 2x^2 - 4x = 0 \implies 2x(x - 2) = 0 \implies x = 0 \text{ or } x = 2.$$ 4. The area between curves is given by $$\int_a^b [f(x) - g(x)] \, dx$$ where $f(x)$ is the upper curve and $g(x)$ the lower. 5. Here, $4x - x^2$ is above $x^2$ on $[0,2]$, so area $$= \int_0^2 [(4x - x^2) - x^2] \, dx = \int_0^2 (4x - 2x^2) \, dx.$$ 6. Compute the integral: $$\int_0^2 4x \, dx = 4 \cdot \frac{x^2}{2} \Big|_0^2 = 4 \cdot 2 = 8,$$ $$\int_0^2 2x^2 \, dx = 2 \cdot \frac{x^3}{3} \Big|_0^2 = 2 \cdot \frac{8}{3} = \frac{16}{3}.$$ 7. So area $$= 8 - \frac{16}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3}.$$ --- 1. Problem 2: Compute the area between $f(x) = e^{-x}$ and $g(x) = e^{-2x}$ on $[0,1]$. 2. Since $e^{-x} > e^{-2x}$ on $[0,1]$, area $$= \int_0^1 (e^{-x} - e^{-2x}) \, dx.$$ 3. Integrate: $$\int e^{-x} dx = -e^{-x} + C,$$ $$\int e^{-2x} dx = -\frac{1}{2} e^{-2x} + C.$$ 4. Evaluate definite integral: $$\int_0^1 (e^{-x} - e^{-2x}) dx = \left[-e^{-x} + \frac{1}{2} e^{-2x}\right]_0^1 = \left(-e^{-1} + \frac{1}{2} e^{-2}\right) - \left(-1 + \frac{1}{2}\right) = 1 - \frac{1}{2} - e^{-1} + \frac{1}{2} e^{-2} = \frac{1}{2} - e^{-1} + \frac{1}{2} e^{-2}.$$ --- 1. Problem 3: Find the area between $y = x^3 - x$ and $y = x - x^2$ over all intersection points. 2. Find intersection points by solving: $$x^3 - x = x - x^2 \implies x^3 - x - x + x^2 = 0 \implies x^3 + x^2 - 2x = 0 \implies x(x^2 + x - 2) = 0.$$ 3. Factor quadratic: $$x^2 + x - 2 = (x+2)(x-1),$$ so roots are $x=0, x=-2, x=1$. 4. Determine which curve is on top in each interval: - For $x \in [-2,0]$, test $x=-1$: $y_1 = (-1)^3 - (-1) = -1 +1=0$, $y_2 = -1 - 1 = -2$, so $y_1 > y_2$. - For $x \in [0,1]$, test $x=0.5$: $y_1 = 0.125 - 0.5 = -0.375$, $y_2 = 0.5 - 0.25 = 0.25$, so $y_2 > y_1$. 5. Area is sum of integrals: $$\int_{-2}^0 (x^3 - x - (x - x^2)) dx + \int_0^1 ((x - x^2) - (x^3 - x)) dx.$$ 6. Simplify integrands: First integral: $$x^3 - x - x + x^2 = x^3 + x^2 - 2x,$$ Second integral: $$x - x^2 - x^3 + x = -x^3 - x^2 + 2x.$$ 7. Compute integrals: $$\int_{-2}^0 (x^3 + x^2 - 2x) dx = \left[\frac{x^4}{4} + \frac{x^3}{3} - x^2\right]_{-2}^0 = (0 + 0 - 0) - \left(\frac{16}{4} - \frac{8}{3} - 4\right) = 0 - (4 - \frac{8}{3} - 4) = \frac{8}{3}.$$ $$\int_0^1 (-x^3 - x^2 + 2x) dx = \left[-\frac{x^4}{4} - \frac{x^3}{3} + x^2\right]_0^1 = \left(-\frac{1}{4} - \frac{1}{3} + 1\right) - 0 = \frac{5}{12}.$$ 8. Total area $$= \frac{8}{3} + \frac{5}{12} = \frac{32}{12} + \frac{5}{12} = \frac{37}{12}.$$ --- 1. Problem 4: Compute area between $x = y^2$ and $x = 2 - y$ using integration w.r.t. $y$. 2. Find intersection points by solving: $$y^2 = 2 - y \implies y^2 + y - 2 = 0,$$ which factors as $$(y+2)(y-1) = 0,$$ so $y = -2$ or $y = 1$. 3. The area is: $$\int_{-2}^1 [(2 - y) - y^2] dy = \int_{-2}^1 (2 - y - y^2) dy.$$ 4. Integrate: $$\int 2 dy = 2y,$$ $$\int y dy = \frac{y^2}{2},$$ $$\int y^2 dy = \frac{y^3}{3}.$$ 5. Evaluate: $$\left[2y - \frac{y^2}{2} - \frac{y^3}{3}\right]_{-2}^1 = \left(2(1) - \frac{1}{2} - \frac{1}{3}\right) - \left(2(-2) - \frac{4}{2} - \frac{-8}{3}\right) = \left(2 - 0.5 - 0.3333\right) - \left(-4 - 2 + 2.6667\right) = 1.1667 - (-3.3333) = 4.5.$$ --- 1. Problem 5: Find the area between $y = \sin(x)$ and $y = \cos(x)$ on $[0, \frac{\pi}{2}]$. 2. Find where $\sin(x) = \cos(x)$: $$\tan(x) = 1 \implies x = \frac{\pi}{4}.$$ 3. On $[0, \frac{\pi}{4}]$, $\cos(x) > \sin(x)$; on $[\frac{\pi}{4}, \frac{\pi}{2}]$, $\sin(x) > \cos(x)$. 4. Area is sum of two integrals: $$\int_0^{\pi/4} (\cos x - \sin x) dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) dx.$$ 5. Integrate: $$\int \cos x dx = \sin x,$$ $$\int \sin x dx = -\cos x.$$ 6. Evaluate first integral: $$\left[\sin x + \cos x\right]_0^{\pi/4} = (\sin \frac{\pi}{4} + \cos \frac{\pi}{4}) - (0 + 1) = \sqrt{2}/2 + \sqrt{2}/2 - 1 = \sqrt{2} - 1.$$ 7. Evaluate second integral: $$\left[-\cos x - \sin x\right]_{\pi/4}^{\pi/2} = \left(-\cos \frac{\pi}{2} - \sin \frac{\pi}{2}\right) - \left(-\cos \frac{\pi}{4} - \sin \frac{\pi}{4}\right) = (-0 - 1) - (-\sqrt{2}/2 - \sqrt{2}/2) = -1 + \sqrt{2}.$$ 8. Total area: $$ (\sqrt{2} - 1) + (-1 + \sqrt{2}) = 2\sqrt{2} - 2 = 2(\sqrt{2} - 1).$$ --- 1. Problem 6: Find the area between $T_1(n) = n^2 + 3n$ and $T_2(n) = 2n^2$. 2. Find intersection points by solving: $$n^2 + 3n = 2n^2 \implies n^2 - 3n = 0 \implies n(n - 3) = 0,$$ so $n=0$ or $n=3$. 3. On $[0,3]$, $2n^2 > n^2 + 3n$ for $n > 0$ (check $n=1$: $2 > 4$ false, so $T_1$ is above $T_2$ on $[0,3]$). 4. Area between curves: $$\int_0^3 [(n^2 + 3n) - 2n^2] dn = \int_0^3 (-n^2 + 3n) dn.$$ 5. Integrate: $$\int -n^2 dn = -\frac{n^3}{3},$$ $$\int 3n dn = \frac{3n^2}{2}.$$ 6. Evaluate: $$\left(-\frac{n^3}{3} + \frac{3n^2}{2}\right)_0^3 = \left(-\frac{27}{3} + \frac{27}{2}\right) - 0 = (-9 + 13.5) = 4.5.$$