Problem: Evaluate the following seven definite integrals. 1. Problem: Compute $$\int_0^1 (3 - 2x)\,dx$$. Formula: Use linearity and the power rule $\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$. Work: Antiderivative is $3x - x^2$. Evaluation: $[3x - x^2]_0^1 = (3\cdot1 - 1^2) - 0 = 2$. Answer: $2$. 2. Problem: Compute $$\int_{-3}^{-2} y(y+1)^2\,dy$$. Formula: Expand and apply the power rule termwise. Work: Expand to $y^3+2y^2+y$ and antiderivative is $\frac{1}{4}y^4 + \frac{2}{3}y^3 + \frac{1}{2}y^2$. Evaluation: $[\frac{1}{4}y^4 + \frac{2}{3}y^3 + \frac{1}{2}y^2]_{-3}^{-2} = (\frac{1}{4}\cdot16 + \frac{2}{3}\cdot(-8) + \frac{1}{2}\cdot4) - (\frac{1}{4}\cdot81 + \frac{2}{3}\cdot(-27) + \frac{1}{2}\cdot9) = \frac{2}{3} - \frac{27}{4} = -\frac{73}{12}$. Answer: $-\frac{73}{12}$. 3. Problem: Compute $$\int_0^a (\sqrt{a} - \sqrt{t})^2\,dt$$. Formula: Expand and use power rule with $\sqrt{t}=t^{1/2}$. Work: Expand to $a - 2\sqrt{a}\,t^{1/2} + t$ and antiderivative is $at - \frac{4}{3}\sqrt{a}\,t^{3/2} + \frac{1}{2}t^2$. Evaluation: At $t=a$ this gives $a^2 - \frac{4}{3}a^2 + \frac{1}{2}a^2 = a^2(1 - \frac{4}{3} + \frac{1}{2}) = \frac{1}{6}a^2$. Answer: $\frac{a^2}{6}$. 4. Problem: Compute $$\int_2^5 (5x-2)(7x+5)\,dx$$. Formula: Multiply and apply power rule termwise. Work: Multiply to get $35x^2+11x-10$ and antiderivative is $\frac{35}{3}x^3 + \frac{11}{2}x^2 -10x$. Evaluation: $[\frac{35}{3}x^3 + \frac{11}{2}x^2 -10x]_2^5 = \frac{4375}{3} + \frac{275}{2} -50 - (\frac{280}{3} + 2) = \frac{2901}{2}$. Answer: $\frac{2901}{2}$. 5. Problem: Compute $$\int_1^3 \frac{m^2 - 2}{m^4}\,dm$$. Formula: Simplify to $m^{-2} - 2m^{-4}$ and use power rule $\int m^n\,dm=\frac{m^{n+1}}{n+1}$. Work: Antiderivative is $-m^{-1} + \frac{2}{3}m^{-3}$. Evaluation: $[-\frac{1}{m} + \frac{2}{3}m^{-3}]_1^3 = (-\frac{1}{3} + \frac{2}{81}) - (-1 + \frac{2}{3}) = \frac{2}{81}$. Answer: $\frac{2}{81}$. 6. Problem: Compute $$\int_{-1/2}^{1/2} (3x+1)^7\,dx$$. Formula: Use substitution $u=3x+1$, $du=3\,dx$. Work: Integral becomes $\frac{1}{3}\int_{-1/2}^{5/2} u^7\,du = \frac{1}{24}[u^8]_{-1/2}^{5/2}$. Evaluation: $\frac{1}{24}\left(\frac{5^8}{2^8} - \frac{1}{2^8}\right) = \frac{390624}{6144} = \frac{4069}{64}$. Answer: $\frac{4069}{64}$. 7. Problem: Compute $$\int_0^2 y^2\sqrt{1+y^3}\,dy$$. Formula: Use substitution $u=1+y^3$, $du=3y^2\,dy$. Work: Integral becomes $\frac{1}{3}\int_1^9 u^{1/2}\,du = \frac{2}{9}[u^{3/2}]_1^9$. Evaluation: $\frac{2}{9}(9^{3/2} - 1) = \frac{2}{9}(27 - 1) = \frac{52}{9}$. Answer: $\frac{52}{9}$.