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
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}$