1. Evaluate $$\int_0^1 x \sin(x^2) \, dx$$ - Use substitution: let $$u = x^2$$, then $$du = 2x \, dx$$ or $$x \, dx = \frac{du}{2}$$. - Change limits: when $$x=0$$, $$u=0$$; when $$x=1$$, $$u=1$$. - Integral becomes $$\int_0^1 \sin(u) \frac{du}{2} = \frac{1}{2} \int_0^1 \sin(u) \, du$$. - Integrate: $$\int \sin(u) \, du = -\cos(u) + C$$. - Evaluate definite integral: $$\frac{1}{2}[-\cos(u)]_0^1 = \frac{1}{2}[-\cos(1) + \cos(0)] = \frac{1}{2}(1 - \cos(1))$$. 2. Evaluate $$\int \frac{5}{(x+3)(x-2)} \, dx$$ - Use partial fractions: $$\frac{5}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}$$. - Multiply both sides by denominator: $$5 = A(x-2) + B(x+3)$$. - Set $$x=2$$: $$5 = A(0) + B(5) \Rightarrow B = 1$$. - Set $$x=-3$$: $$5 = A(-5) + B(0) \Rightarrow A = -1$$. - Integral becomes $$\int \left( \frac{-1}{x+3} + \frac{1}{x-2} \right) dx = -\ln|x+3| + \ln|x-2| + C = \ln\left| \frac{x-2}{x+3} \right| + C$$. 3. Evaluate $$\int_0^1 \int_0^z \int_0^y x \, dx \, dy \, dz$$ - Inner integral: $$\int_0^y x \, dx = \left[ \frac{x^2}{2} \right]_0^y = \frac{y^2}{2}$$. - Middle integral: $$\int_0^z \frac{y^2}{2} \, dy = \frac{1}{2} \left[ \frac{y^3}{3} \right]_0^z = \frac{z^3}{6}$$. - Outer integral: $$\int_0^1 \frac{z^3}{6} \, dz = \frac{1}{6} \left[ \frac{z^4}{4} \right]_0^1 = \frac{1}{24}$$. 4. Evaluate $$\int \frac{x+1}{\sqrt{x}} \, dx$$ - Rewrite integrand: $$\frac{x}{\sqrt{x}} + \frac{1}{\sqrt{x}} = x^{1/2} + x^{-1/2}$$. - Integrate termwise: $$\int x^{1/2} \, dx = \frac{2}{3} x^{3/2} + C$$ and $$\int x^{-1/2} \, dx = 2 x^{1/2} + C$$. - Sum: $$\frac{2}{3} x^{3/2} + 2 x^{1/2} + C$$. 5. Evaluate $$\int \frac{1}{16 + x^2} \, dx$$ - Use formula: $$\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \arctan\left( \frac{x}{a} \right) + C$$. - Here, $$a=4$$. - Result: $$\frac{1}{4} \arctan\left( \frac{x}{4} \right) + C$$. 6. Evaluate $$\int x \sin x \, dx$$ - Use integration by parts: let $$u = x$$, $$dv = \sin x \, dx$$. - Then $$du = dx$$, $$v = -\cos x$$. - Integral: $$uv - \int v \, du = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x + C$$. 7. Find area bounded by $$x = y^2$$ and $$x = 8 - y^2$$ - Set equal: $$y^2 = 8 - y^2 \Rightarrow 2 y^2 = 8 \Rightarrow y^2 = 4 \Rightarrow y = \pm 2$$. - Area between curves: $$\int_{-2}^2 [(8 - y^2) - y^2] \, dy = \int_{-2}^2 (8 - 2 y^2) \, dy$$. - Integrate: $$\left[ 8y - \frac{2 y^3}{3} \right]_{-2}^2 = \left(16 - \frac{16}{3}\right) - \left(-16 + \frac{16}{3}\right) = 32 - \frac{32}{3} = \frac{64}{3}$$. 8. Find displacement from $$t=1$$ to $$t=2$$ for velocity $$v = \frac{t+1}{(t^2 + 2t)^2}$$ - Displacement: $$\int_1^2 \frac{t+1}{(t^2 + 2t)^2} \, dt$$. - Note $$t^2 + 2t = t(t+2)$$. - Let $$u = t^2 + 2t$$, then $$du = (2t + 2) dt = 2(t+1) dt$$. - So $$dt = \frac{du}{2(t+1)}$$. - Substitute: $$\int \frac{t+1}{u^2} dt = \int \frac{t+1}{u^2} \cdot \frac{du}{2(t+1)} = \frac{1}{2} \int u^{-2} du = \frac{1}{2} \left(-u^{-1}\right) + C = -\frac{1}{2u} + C$$. - Evaluate definite integral: $$\left[-\frac{1}{2(t^2 + 2t)}\right]_1^2 = -\frac{1}{2(4 + 4)} + \frac{1}{2(1 + 2)} = -\frac{1}{16} + \frac{1}{6} = \frac{5}{48}$$. Final answers: 1. $$\frac{1}{2}(1 - \cos(1))$$ 2. $$\ln\left| \frac{x-2}{x+3} \right| + C$$ 3. $$\frac{1}{24}$$ 4. $$\frac{2}{3} x^{3/2} + 2 x^{1/2} + C$$ 5. $$\frac{1}{4} \arctan\left( \frac{x}{4} \right) + C$$ 6. $$-x \cos x + \sin x + C$$ 7. $$\frac{64}{3}$$ 8. $$\frac{5}{48}$$