constant):
Z
audxDa
Z
udx (4.46)Z
.uCv/dxDZ
udxCZ
vdx (4.47)Z
.uv/dxDZ
udxZ
vdx (4.48)These results will be proved in a more rigorous calculus course. There are no product or quotient rules for
integrals as there are for derivatives.
Since the derivative and integration are inverses of each other, and the functionexis equal to its own
derivative, it is also equal to its own integral (to within an arbitrary constant of integration):
Z
exdxDexCC (4.49)
Example.Find the indefinite integral of the functionf.x/D4x^3 C7x^2 5xC 6 with respect tox, and
find the area underf.x/betweenxD 3 andxD 4.
Solution.Using the above results,
Z
f.x/dxD
Z
.4x^3 C7x^2 5xC6/dx (4.50)D
Z
4x^3 dxCZ
7x^2 dxZ
5x dxCZ
6dx (4.51)D 4
Z
x^3 dxC 7Z
x^2 dx 5Z
xdxC 6Z
dx (4.52)D 4
x^4
4CC 1 C 7
x^3
3CC 2 5
x^2
2CC 3 C6.x/CC 4 (4.53)Dx^4 C7
3
x^3 5
2
x^2 C6xCC (4.54)where we have combined all the individual constants of integrationC 1 ,C 2 ,C 3 ,C 4 into a single constantC.
To find the area under the curve betweenxD 3 andxD 4 , we compute the definite integral
Z 4
3f.x/dx (4.55)We’ve already found the indefinite integral; all we need to do is evaluate the indefinite integral atxD 4 , and
subtract the indefinite integral evaluated atxD 3 :
areaDZ 4
3f.x/dxDx^4 C7
3
x^3 5
2
x^2 C6xCCˇˇ
ˇ
ˇ
ˇ
43