Question #277981

 Evaluate \intopx2(1 + 2x3)3dx.


2. Evaluate \intopxe7x 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].



1
Expert's answer
2021-12-16T12:18:30-0500

(1)  I=x2(1+2x3)3dx  By  substitution  method,  let  u=1+2x3Such  that  du  =  6x2  dx   I=x2(1+2x3)3dx  =I=\x2u3du6\x2  I=14u3du  =  u424+CI=(1+2x3)424+C(2)  I =x e7x  dxUsing  integration  by  part   u=x   ,  dv=e7x  dxdu=1  ,  v=e7x  7udv=uv    v  du I=xe7x  717 e7x  dxI=xe7x  7e7x49+C(3)  y=1+x2Using  method  of  diskV  =  22πy2dx   =  22π(1+x2)2dxV  =  22π(1+2x2+x4)dxV=412π15\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}}


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