Calculus II - Applications of integration

19 flashcards · Mathematics · by luckydog789 (harrison holbrook)

1. Area bounded by two graphs

   The area AAA bounded by the graphs of y=f(x)y=f(x)y=f(x) and y=g(x)y=g(x)y=g(x) , and the lines x=ax=ax=a and x=bx=bx=b , where fff and ggg are continuous on the interval [a,b][a,b][a,b] and f(x)≥g(x)f(x) \geq g(x)f(x)≥g(x) for all numbers xxx in [a,b][a,b][a,b] is A=∫ab[f(x)−g(x)]dxA=\int_a^b[f(x)-g(x)]dxA=∫ab​[f(x)−g(x)]dx

2. Area bounded by two graphs by partitioning the y-axis

   The area AAA bounded by the graphs of x=f(y)x=f(y)x=f(y) and x=g(y)x=g(y)x=g(y) , and the horizontal lines y=cy=cy=c and y=dy=dy=d , where fff and ggg are continuous on the interval [c,d][c,d][c,d] and f(y)≥g(y)f(y) \geq g(y)f(y)≥g(y) for all numbers yyy in [c,d][c,d][c,d] is A=∫cd[f(y)−g(y)]dyA=\int_c^d[f(y)-g(y)]dyA=∫cd​[f(y)−g(y)]dy (To find x=f(y)x=f(y)x=f(y) , solve for xxx in y=f(x)y=f(x)y=f(x) )

3. Volume of a disk

   V=(Cross Section)(Thickness)=πr2hV=(\text{Cross Section})(\text{Thickness})=\pi r^2hV=(Cross Section)(Thickness)=πr2h

4. volume of a solid of revolution using the disk method

   If a function fff is continuous and nonnegative on a closed interval [a,b][a,b][a,b] , then the volume VVV of the solid of revolution formed by revolving the region bounded by the graph of yyy , the xxx -axis, and the lines x=ax=ax=a and x=bx=bx=b about the xxx -axis is V=π∫ab[f(x)]2dxV=\pi\int_a^b[f(x)]^2dxV=π∫ab​[f(x)]2dx (This is a special case of the washer method)

5. Volume of a washer

   V=π[(Outer)2−(Inner)2](Thickness)V=\pi[(\text{Outer})^2-(\text{Inner})^2](\text{Thickness})V=π[(Outer)2−(Inner)2](Thickness)

6. Volume of a solid of revolution using the washer method

   If the functions y=f(x)y=f(x)y=f(x) and y=g(x)y=g(x)y=g(x) are continuous on the closed interval [a,b][a,b][a,b] and if f(x)≥g(x)≥0f(x) \geq g(x) \geq 0f(x)≥g(x)≥0 on [a,b][a,b][a,b] , then the volume VVV of a solid of revolution obtained by revolving the region bounded by the graphs of fff and ggg and the lines x=ax=ax=a and x=bx=bx=b about the xxx -axis is V=π∫ab([f(x)]2−[g(x)2])dxV=\pi\int_a^b([f(x)]^2-[g(x)^2])dxV=π∫ab​([f(x)]2−[g(x)2])dx

7. Volume of a cylindrical shell

   V=(Circumference)(Height)(Thickness)=2πrh(Thickness)V=(\text{Circumference})(\text{Height})(\text{Thickness})=2\pi rh(\text{Thickness})V=(Circumference)(Height)(Thickness)=2πrh(Thickness)

8. Volume of a solid of revolution about the y-axis via the shell method

   If y=f(x)y=f(x)y=f(x) is a function that is continuous and nonnegative on the closed interval [a,b][a,b][a,b] , where a≥0a \geq 0a≥0 , then the volume VVV of the solid generated by revolving the region bounded by the graph of fff , the xxx -axis, and the lines x=ax=ax=a and x=bx=bx=b about the yyy -axis is V=2π∫abxf(x)dxV=2\pi\int_a^bxf(x)dxV=2π∫ab​xf(x)dx

9. Volume of a solid of revolution about the x-axis via the shell method

   If x=g(y)x=g(y)x=g(y) is a function that is continuous and nonnegative on the closed interval [c,d][c,d][c,d] , where c≥0c \geq 0c≥0 , then the volume VVV of the solid generated by revolving the region bounded by the graph of x=g(y)x=g(y)x=g(y) , the yyy -axis, and the lines y=cy=cy=c and y=dy=dy=d about the xxx -axis is V=2π∫cdyg(y)dyV=2\pi\int_c^dyg(y)dyV=2π∫cd​yg(y)dy

10. Summary of solids of revolution (for about the yyy -axis just change every xxx to yyy and every yyy to xxx )

11. Volume of a slice

    Vi=(Cross section)(Thickness)=A(ui)ΔxV_i=(\text{Cross section})(\text{Thickness})=A(u_i)\Delta xVi​=(Cross section)(Thickness)=A(ui​)Δx

12. Volume of a solid using slicing

    If for each xxx in [a,b][a,b][a,b] , the area A(x)A(x)A(x) of the cross sections of a solid is known and is continuous on [a,b][a,b][a,b] , then the volume VVV of the solid is V=∫abA(x)dxV=\int_a^bA(x)dxV=∫ab​A(x)dx To find volume, first find a function for the area then solve using the formula above

13. Arc length

    Suppose a function y=f(x)y=f(x)y=f(x) is continuous on [a,b][a,b][a,b] and has a derivative that is continuous on an interval containing aaa and bbb . The arc length sss of the graph of fff from x=ax=ax=a to x=bx=bx=b is given by s=∫ab1+[f′(x)]2dxs=\int_a^b \sqrt{1+[f'(x)]^2}dxs=∫ab​1+[f′(x)]2 ​dx

14. Arc length (partitioning the y-axis)

    Suppose a function x=g(y)x=g(y)x=g(y) is continuous on [c,d][c,d][c,d] and has a derivative that is continuous on an interval containing ccc and ddd . The arc length sss of the graph of ggg from y=cy=cy=c to y=dy=dy=d is given by s=∫cd1+[g′(y)]2dys=\int_c^d \sqrt{1+[g'(y)]^2}dys=∫cd​1+[g′(y)]2 ​dy

15. Work done by a constant force

    W=FxW=FxW=Fx , where WWW is work in Joules, FFF is force in Newtons, and xxx is displacement in meters

16. Work done by a variable force

    The work WWW done by a continuously varying force FFF acting on an object, which moves the object along a straight line in the direction FFF from x=ax=ax=a to x=bx=bx=b is W=∫abF(x)dxW=\int_a^bF(x)dxW=∫ab​F(x)dx

17. Hooke's law

    The spring force, F(x)F(x)F(x) , in Newtons, is equal to F(x)=−kxF(x)=-kxF(x)=−kx , where kkk is the sprint constant for the spring and xxx is the displacement of the spring from its equilibrium.

18. Work to lift an object

    W=(Weight)(Distance lifted)=mg(Distance lifted)W=(\text{Weight})(\text{Distance lifted})=mg(\text{Distance lifted})W=(Weight)(Distance lifted)=mg(Distance lifted)

19. Work to move an object ddd meters above the surface of Earth

    W=gRm(1−RR+d),R=6.37⋅106mW=gRm(1-\frac{R}{R+d}), R=6.37\cdot10^6 \text{m}W=gRm(1−R+dR​),R=6.37⋅106m