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EXAMPLE 4Using Part 2 of the Fundamental Theorem of Calculus

Use Part 2 of the Fundamental Theorem of Calculus to find:

  1. (a) 12x2dx
  2. (b) π/60cosxdx
  3. (c) 3/2011x2dx
  4. (d) 211xdx

Solution (a) An antiderivative of f(x)=x2 is F(x)=x33. By Part 2 of the Fundamental Theorem of Calculus, 12x2dx=[x33]12=133(2)33=13+83=93=3

(b) An antiderivative of f(x)=cosx is F(x)=sinx. By Part 2 of the Fundamental Theorem of Calculus, π/60cosxdx=[sinx]π/60=sinπ6sin0=12

(c) An antiderivative of f(x)=11x2 is F(x)=sin1x, provided |x|<1. By Part 2 of the Fundamental Theorem of Calculus, 3/2011x2dx=[sin1x]3/20=sin132sin10=π30=π3

(d) An antiderivative of f(x)=1x is F(x)=ln|x|. By Part 2 of the Fundamental Theorem of Calculus, 211xdx=[ln|x|]21=ln2ln1=ln20=ln2