# Thread: fundamental theorem of calculus

1. ## fundamental theorem of calculus

fundamental theorem of calculus

find an expression for g(x)= integral from 0 to x of f(x)dx similar to the one for f(x).

f(x) = x+3 for x less than or equal to -1
f(x) = x^2 +1 for x > -1

g(x)= integral from 0 to x of f(x)dx

2. Originally Posted by cm3pyro
fundamental theorem of calculus

find an expression for g(x)= integral from 0 to x of f(x)dx similar to the one for f(x).

f(x) = x+3 for x less than or equal to -1
f(x) = x^2 +1 for x > -1

g(x)= integral from 0 to x of f(x)dx

Looks straight forward. If $\displaystyle -1\le x$, g(x)= $\displaystyle \int_0^x f(t)dt= \int t^2+ 1 dt= \frac{x^3}{3}+ x$. If x< -1 g(x)= $\displaystyle \int_0^x f(t)dt$$\displaystyle = \int_0^-1 t^2+ 1 dt+ \int_{-1}^x t+ 3 dt$$\displaystyle = -\frac{4}{3}+ \int_{-1}^x t+ 3 dt$$\displaystyle = -4/3+ \frac{x^2}{2}+ 3x- \frac{7}{2}$

3. for the second function, x>-1

you said, x<-1

will this change my answer?

4. find an expression for g(x)= $\displaystyle \int_0^x$ f(x)dx similar to the one for f(x).

f(x) = x+3 for x$\displaystyle \le$ -1
f(x) = x^2 +1 for x > -1

g(x)= $\displaystyle \int_0^x$ f(x)dx

5. still need help with this problem, the above poster was incorrect. He switched up the functions and the bounds.. please help