Answer to Question #277981 in Calculus for Sam

Question

 Evaluate $\intop$x²(1 + 2x³)³dx.

2. Evaluate $\intop$xe^7x dx.

3. Find the volume of the solid of revolution when the curve y = 1 + x2 is revolved around the x-axis on [−2, 2].

Answer 1

$\left(\mathrm{1}\right)\ \ I=\int{x^{\mathrm{2}}{\left(\mathrm{1}+\mathrm{2}x^{\mathrm{3}}\right)}^{\mathrm{3}}dx\ \ } \\ By\ \ substitution\ \ method,\ \ let\ \ u=\mathrm{1}+\mathrm{2}x^{\mathrm{3}} \\ Such\ \ that\ \ du\ \ =\ \ \mathrm{6}x^{\mathrm{2}}\ \ dx\ \ \ \\ I=\int{x^{\mathrm{2}}{\left(\mathrm{1}+\mathrm{2}x^{\mathrm{3}}\right)}^{\mathrm{3}}dx\ \ }=I=\int{\rlap{\textbackslash}x^{\mathrm{2}}u^{\mathrm{3}}\frac{du}{\mathrm{6}\rlap{\textbackslash}x^{\mathrm{2}}}\ \ } \\ I=\frac{\mathrm{1}}{\mathrm{4}}\int{u^{\mathrm{3}}du\ \ }=\ \ \frac{u^{\mathrm{4}}}{\mathrm{24}}+C \\ I=\frac{{\left(\mathrm{1}+\mathrm{2}x^{\mathrm{3}}\right)}^{\mathrm{4}}}{\mathrm{24}}+C \\ \\ \left(\mathrm{2}\right)\ \ I\ =\int{x\ e^{\mathrm{7}x}\ \ dx} \\ U\mathrm{sin}g\ \ \mathrm{int}egration\ \ by\ \ part\ \ \ \\ u=x\ \ \ ,\ \ dv=e^{\mathrm{7}x}\ \ dx \\ du=\mathrm{1}\ \ ,\ \ v=\frac{e^{\mathrm{7}x}\ \ }{\mathrm{7}} \\ \int{udv}=uv\ \ -\ \ \int{v\ \ du\ } \\ \\ I=\frac{xe^{\mathrm{7}x}\ \ }{\mathrm{7}}-\frac{\mathrm{1}}{\mathrm{7}}\int{\ e^{\mathrm{7}x}\ \ dx} \\ \\ I=\frac{xe^{\mathrm{7}x}\ \ }{\mathrm{7}}-\frac{e^{\mathrm{7}x}}{\mathrm{49}}+C \\ \\ \left(\mathrm{3}\right)\ \ y=\mathrm{1}+x^{\mathrm{2}} \\ U\mathrm{sin}g\ \ method\ \ of\ \ disk \\ V\ \ =\ \ \int^{\mathrm{2}}_{-\mathrm{2}}{\pi y^{\mathrm{2}}dx\ \ \ =\ \ }\int^{\mathrm{2}}_{-\mathrm{2}}{\pi {\left(\mathrm{1}+x^{\mathrm{2}}\right)}^{\mathrm{2}}dx} \\ \\ V\ \ =\ \ \int^{\mathrm{2}}_{-\mathrm{2}}{\pi \left(\mathrm{1}+\mathrm{2}x^{\mathrm{2}}+x^{\mathrm{4}}\right)dx} \\ \\ V=\frac{\mathrm{412}\pi }{\mathrm{15}}$