1. Evaluate the double integral $$\int_0^1 \int_y^1 e^{-x^2} \, dx \, dy$$. - The integral is over the region where $y$ goes from 0 to 1 and $x$ goes from $y$ to 1. - Since $e^{-x^2}$ has no elementary antiderivative, consider reversing the order of integration. 2. Reverse the order of integration: - Original limits: $0 \le y \le 1$, $y \le x \le 1$. - New limits: $0 \le x \le 1$, $0 \le y \le x$. 3. Rewrite the integral: $$\int_0^1 \int_0^x e^{-x^2} \, dy \, dx$$ - Since $e^{-x^2}$ does not depend on $y$, integrate with respect to $y$ first: $$\int_0^1 e^{-x^2} (x - 0) \, dx = \int_0^1 x e^{-x^2} \, dx$$ 4. Use substitution to evaluate: - Let $u = -x^2$, then $du = -2x \, dx$ or $-\frac{1}{2} du = x \, dx$. - Change limits: when $x=0$, $u=0$; when $x=1$, $u=-1$. - Integral becomes: $$\int_0^1 x e^{-x^2} \, dx = -\frac{1}{2} \int_0^{-1} e^{u} \, du = -\frac{1}{2} [e^{u}]_0^{-1} = -\frac{1}{2} (e^{-1} - 1) = \frac{1}{2} (1 - e^{-1})$$ --- 5. Find the area enclosed by the lines: $$y = -x, \quad -x + y = 2, \quad y = 2$$ - Rewrite $-x + y = 2$ as $y = x + 2$. - The three lines intersect forming a triangle. 6. Find intersection points: - Intersection of $y = -x$ and $y = 2$: set $-x = 2 \Rightarrow x = -2$, point $(-2, 2)$. - Intersection of $y = x + 2$ and $y = 2$: set $x + 2 = 2 \Rightarrow x = 0$, point $(0, 2)$. - Intersection of $y = -x$ and $y = x + 2$: set $-x = x + 2 \Rightarrow 2x = -2 \Rightarrow x = -1$, then $y = -(-1) = 1$, point $(-1, 1)$. 7. Set up double integral for area: - The region is bounded between $x = -2$ and $x = 0$. - For $x$ in $[-2, -1]$, $y$ goes from $y = -x$ to $y = 2$. - For $x$ in $[-1, 0]$, $y$ goes from $y = x + 2$ to $y = 2$. - Area = $$\int_{-2}^{-1} \int_{-x}^2 dy \, dx + \int_{-1}^0 \int_{x+2}^2 dy \, dx$$ 8. Evaluate integrals: - First integral: $$\int_{-2}^{-1} (2 - (-x)) \, dx = \int_{-2}^{-1} (2 + x) \, dx = [2x + \frac{x^2}{2}]_{-2}^{-1} = ( -2 + \frac{1}{2} ) - ( -4 + 2 ) = (-1.5) - (-2) = 0.5$$ - Second integral: $$\int_{-1}^0 (2 - (x + 2)) \, dx = \int_{-1}^0 (-x) \, dx = [-\frac{x^2}{2}]_{-1}^0 = 0 - (-\frac{1}{2}) = \frac{1}{2}$$ - Total area = $0.5 + 0.5 = 1$. --- 9. Find volume bounded above by plane $x + y + z = 3$ and below by rectangle $x = -2$, $x = 0$, $y = 1$, $y = 5$. - Solve for $z$: $z = 3 - x - y$. - Volume = $$\int_{x=-2}^0 \int_{y=1}^5 (3 - x - y) \, dy \, dx$$ 10. Evaluate inner integral: $$\int_1^5 (3 - x - y) \, dy = [3y - xy - \frac{y^2}{2}]_1^5 = (15 - 5x - \frac{25}{2}) - (3 - x - \frac{1}{2}) = (15 - 5x - 12.5) - (3 - x - 0.5) = (2.5 - 5x) - (2.5 - x) = 2.5 - 5x - 2.5 + x = -4x$$ 11. Now integrate with respect to $x$: $$\int_{-2}^0 -4x \, dx = [-2x^2]_{-2}^0 = 0 - (-8) = 8$$ - Volume = 8. --- 12. Find area of region $R$ enclosed by: $$y = -x + 1, \quad y = e^x, \quad y = 0, \quad x = 2$$ - Find intersection of $y = -x + 1$ and $y = e^x$. - Numerically, they intersect near $x=0.5$. 13. Set up integral for area: - For $x$ in $[0, 0.5]$, upper curve is $y = -x + 1$. - For $x$ in $[0.5, 2]$, upper curve is $y = e^x$. - Lower curve is $y=0$. - Area = $$\int_0^{0.5} (-x + 1) \, dx + \int_{0.5}^2 e^x \, dx$$ 14. Evaluate integrals: - $$\int_0^{0.5} (-x + 1) \, dx = \left[-\frac{x^2}{2} + x\right]_0^{0.5} = \left(-\frac{0.25}{2} + 0.5\right) - 0 = ( -0.125 + 0.5 ) = 0.375$$ - $$\int_{0.5}^2 e^x \, dx = [e^x]_{0.5}^2 = e^2 - e^{0.5}$$ - Total area = $$0.375 + e^2 - e^{0.5}$$ --- 15. Find volume bounded by $f(x,y) = 2xy$ and region $R$ in first quadrant between parabola $y^2 = 4x$ and line $y = \frac{1}{2} x$. - Express $x$ in terms of $y$ for parabola: $x = \frac{y^2}{4}$. - Find intersection points by equating $y = \frac{1}{2} x$ and $y^2 = 4x$. - Substitute $x = 2y$ into parabola: $$y^2 = 4(2y) = 8y \Rightarrow y^2 - 8y = 0 \Rightarrow y(y - 8) = 0$$ - So $y=0$ or $y=8$. - For $y$ in $[0,8]$, $x$ goes from parabola $x = \frac{y^2}{4}$ to line $x = 2y$. 16. Set up volume integral: $$\int_0^8 \int_{\frac{y^2}{4}}^{2y} 2xy \, dx \, dy$$ 17. Evaluate inner integral: $$\int_{\frac{y^2}{4}}^{2y} 2xy \, dx = 2y \int_{\frac{y^2}{4}}^{2y} x \, dx = 2y \left[ \frac{x^2}{2} \right]_{\frac{y^2}{4}}^{2y} = 2y \left( \frac{(2y)^2}{2} - \frac{(\frac{y^2}{4})^2}{2} \right) = 2y \left( \frac{4y^2}{2} - \frac{y^4/16}{2} \right) = 2y \left( 2y^2 - \frac{y^4}{32} \right) = 4y^3 - \frac{y^5}{16}$$ 18. Now integrate with respect to $y$: $$\int_0^8 \left(4y^3 - \frac{y^5}{16} \right) dy = \left[ y^4 - \frac{y^6}{96} \right]_0^8 = 8^4 - \frac{8^6}{96} = 4096 - \frac{262144}{96} = 4096 - 2730.6667 = 1365.3333$$ - Volume = $\frac{4096 \times 96 - 262144}{96} = 1365 \frac{1}{3}$ approximately. --- Final answers: 1. $$\frac{1}{2} (1 - e^{-1})$$ 2. Area = 1 3. Volume = 8 4. Area = $$0.375 + e^2 - e^{0.5}$$ 5. Volume = $$1365 \frac{1}{3}$$ (or exactly $4096 - \frac{262144}{96}$)