Calculus, Chapter 6

G(x) = F(x) + C

C is the constant of integration

y = f(x)dx = F(x) + C

Given: (3x²-1)dx = x³-x+C Passes through (2,4)

F(x) = x³-x+C F(2) = 8-2+C F(2) = 4 when C= -2

So, particular solution: F(x) = x³-x-2

• n
• a_i = a₁ + a₂ + a₃ +...+a_n
• i=1

Sigma i = [n(n+1)]/2

Sigma i² = [n(n+1)(2n+1)]/6

Sigma i³ = [n²(n+1)²]/4

s(n) = Sigma f(m₁)(change x)

m_i = 0 + (i-1)(change x)

S(n) = Sigma f(M₁)(change x)

M_i = 0+i(change x)

the area of a region bounded by the graph of f, the x-axis, and the vertical lines x=a and x=b is:

Area = _limSigma f(c_i)(change x) where (change x)=[b-a]/n

• b
• Sigma f(x)dx = _lim Sigma f(c_i)(change x_i) = F(b)-F(a) = F(x)
• a

a=lower limit; b=upper limit

*remember the line thing...

• b
• [1/(b-a)]Sigma f(x)dx = f(C)
• a

if f is continuous on the closed interval [a,b] then there exixts a number c in the closed interval [a,b] such that:

• b
• Sigma f(x)dx = f(c)(b-a)
• a

When we defined the definite integral of f on the interval [a,b] we used the constant b as the upper limit of integration and x as the variable of integration. We now look at a slightly different situation in which the variable x is used as the upper limit of integration.

(d/dx)Sigma dx

• 1. Choose a Substitution; choose the inner part of a composite function to sub
• 2. Compute du = g'(x)dx
• 3. Rewrite the integral in terms of the variable u
• 4. Evaluate the resulting integral in terms of u
• 5. Replace u by g(x) to obtain an antiderivative in terms of x
• 6. Check your answer by differentiating

Sigma [g(x)]ⁿ g'(x)dx = {[[g(x)']ⁿ⁺¹]/n+1} +C

• Equivalently, if u=g(x) then:
• Sigma uⁿdu = [(uⁿ⁺¹)/n+1]+C