1. **Problem 1:** Evaluate the definite integral $$\int_1^0 (3x+2)^4 \, dx$$. - We use the substitution method. Let $$u = 3x + 2$$, then $$du = 3 \, dx$$ or $$dx = \frac{du}{3}$$. - When $$x=1$$, $$u=3(1)+2=5$$; when $$x=0$$, $$u=3(0)+2=2$$. - The integral becomes $$\int_5^2 u^4 \frac{du}{3} = \frac{1}{3} \int_5^2 u^4 \, du$$. - Reverse the limits to get rid of the negative sign: $$-\frac{1}{3} \int_2^5 u^4 \, du$$. - Integrate: $$\int u^4 \, du = \frac{u^5}{5}$$. - So the integral is $$-\frac{1}{3} \left[ \frac{u^5}{5} \right]_2^5 = -\frac{1}{15} (5^5 - 2^5) = -\frac{1}{15} (3125 - 32) = -\frac{3093}{15} = -206.2$$. 2. **Problem 2:** Evaluate the definite integral $$\int_0^1 \frac{2x}{x^2+4} \, dx$$. - Use substitution: let $$u = x^2 + 4$$, then $$du = 2x \, dx$$. - When $$x=0$$, $$u=0^2+4=4$$; when $$x=1$$, $$u=1^2+4=5$$. - The integral becomes $$\int_4^5 \frac{1}{u} \, du$$. - Integrate: $$\int \frac{1}{u} \, du = \ln|u|$$. - Evaluate: $$[\ln u]_4^5 = \ln 5 - \ln 4 = \ln \frac{5}{4}$$. 3. **Problem 3:** Evaluate the definite integral $$\int_0^1 \frac{x^3}{1+x^2} \, dx$$. - Rewrite numerator: $$x^3 = x \cdot x^2$$. - Split the integral: $$\int_0^1 \frac{x^3}{1+x^2} \, dx = \int_0^1 \frac{x(x^2+1-1)}{1+x^2} \, dx = \int_0^1 x \, dx - \int_0^1 \frac{x}{1+x^2} \, dx$$. - First integral: $$\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}$$. - Second integral: use substitution $$u=1+x^2$$, $$du=2x \, dx$$, so $$x \, dx = \frac{du}{2}$$. - When $$x=0$$, $$u=1$$; when $$x=1$$, $$u=2$$. - The second integral becomes $$\frac{1}{2} \int_1^2 \frac{1}{u} \, du = \frac{1}{2} [\ln u]_1^2 = \frac{1}{2} \ln 2$$. - Final answer: $$\frac{1}{2} - \frac{1}{2} \ln 2 = \frac{1}{2} (1 - \ln 2)$$. **Summary of answers:** 1. $$-206.2$$ 2. $$\ln \frac{5}{4}$$ 3. $$\frac{1}{2} (1 - \ln 2)$$