Integration rule for piecewise definition by interval

Statement

Definite integration version with fixed endpoints

The process is as follows:

• Break up the interval of integration into pieces based on the pieces of definition of the function.
• Compute the definite integral on each piece of the corresponding function.
• Add them up.

Definite integral version with fixed lower endpoint, variable upper endpoint

The process is as follows:

• Make cases for the variable upper endpoint based on which interval it lands inside.
• For each case, use the "definite integration version with fixed endpoints"

Indefinite integral computation

Simply do the "definite integral version with fixed lower endpoint, variable upper endpoint" by making an arbitrary choice of fixed lower endpoint (it is usually most convenient to choose this as the left endpoint of the interval of definition of the function if such a point exists). After getting the answer, add a $+C$ to it (or rather, to each of the pieces).

Examples

Definite integral version with fixed endpoints

Consider the function:

$f(x):=\left\lbrace {\begin{array}{rr}-5,&x\leq -5\\x,&-5<x\leq 0\\x^{2},&0<x<1\\x^{3},&x\geq 1\\\end{array}}\right.$

Suppose we are asked to calculate:

$\int _{-1}^{3}f(x)\,dx$

We first break up the interval of integration into pieces based on the function definition. We note that the only points in $[-1,3]$ where the function definition changes are the points 0 and 1, so we break up the interval of integration at 0 and 1. We thus get three intervals: $[-1,0],[0,1],and[1,3]$. Thus:

$\int _{-1}^{3}f(x)\,dx=\int _{-1}^{0}f(x)\,dx+\int _{0}^{1}f(x)\,dx+\int _{1}^{3}f(x)\,dx$

Within each interval, we can use the definition for that interval:

$\int _{-1}^{3}f(x)\,dx=\int _{-1}^{0}x\,dx+\int _{0}^{1}x^{2}\,dx+\int _{1}^{3}x^{3}\,dx$

This becomes:

$\int _{-1}^{3}f(x)\,dx=\left[{\frac {x^{2}}{2}}\right]_{-1}^{0}+\left[{\frac {x^{3}}{3}}\right]_{0}^{1}+\left[{\frac {x^{4}}{4}}\right]_{1}^{3}$

This simplifies to:

$\int _{-1}^{3}f(x)\,dx={\frac {-1}{2}}+{\frac {1}{3}}+20={\frac {119}{6}}$

Definite integral version with fixed lower endpoint, variable upper endpoint; also tackles indefinite integral

Consider the function:

$f(x):=\left\lbrace {\begin{array}{rr}x^{5},&x<-1\\x,&-1\leq x\leq 0\\x^{2},&0<x<1\\x^{3},&x\geq 1\\\end{array}}\right.$

Suppose we are asked to determine the following as a function of $x$:

$\int _{0}^{x}f(t)\,dt$

We make cases based on the interval in which $x$ lies.

Suppose $x<-1$. The interval of integration breaks at -1, and we get:

$\int _{0}^{x}f(t)\,dt=\int _{-1}^{x}f(t)\,dt+\int _{0}^{-1}f(t)\,dt=\int _{-1}^{x}t^{5}\,dt+\int _{0}^{-1}t\,dt=\left[{\frac {t^{6}}{6}}\right]_{-1}^{x}+\left[{\frac {t^{2}}{2}}\right]_{0}^{-1}={\frac {x^{6}-1}{6}}+{\frac {1}{2}}={\frac {1}{3}}+{\frac {x^{6}}{6}}$

Suppose now that $-1\leq x\leq 0$. In this case, we can just do the integration directly, since the function has a single definition. We get:

$\int _{0}^{x}f(t)\,dt=\int _{0}^{x}t\,dt=\left[{\frac {t^{2}}{2}}\right]_{0}^{x}={\frac {x^{2}}{2}}$

Suppose now that $0<x<1$. In this case, we can again do the integration directly, and we get:

$\int _{0}^{x}f(t)\,dt=\int _{0}^{x}t^{2}\,dt=\left[{\frac {t^{3}}{3}}\right]_{0}^{x}={\frac {x^{3}}{3}}$

Finally, consider the case that $x\geq 1$. In this case, the interval of integration must be split at 1 where the function definition changes, so we get:

$\int _{0}^{x}f(t)\,dt=\int _{0}^{1}t^{2}\,dt+\int _{1}^{x}t^{3}\,dt=\left[{\frac {t^{3}}{3}}\right]_{0}^{1}+\left[{\frac {t^{4}}{4}}\right]_{1}^{x}={\frac {1}{3}}+{\frac {x^{4}-1}{4}}={\frac {x^{4}}{4}}+{\frac {1}{12}}$

The overall definition is:

$\int _{0}^{x}f(t)\,dt=\left\lbrace {\begin{array}{rr}{\frac {1}{3}}+{\frac {x^{6}}{6}},&x<-1\\{\frac {x^{2}}{2}},&-1\leq x\leq 0\\{\frac {x^{3}}{3}},&0<x<1\\{\frac {x^{4}}{4}}+{\frac {1}{12}},&x\geq 1\\\end{array}}\right.$

The indefinite integral is:

$\int f(x)\,dx=\left\lbrace {\begin{array}{rr}{\frac {1}{3}}+{\frac {x^{6}}{6}}+C,&x<-1\\{\frac {x^{2}}{2}}+C,&-1\leq x\leq 0\\{\frac {x^{3}}{3}}+C,&0<x<1\\{\frac {x^{4}}{4}}+{\frac {1}{12}}+C,&x\geq 1\\\end{array}}\right.,\qquad C\in \mathbb {R}$