# Video: Evaluating the Integral of a Piecewise Function

For the piecewise function π(π₯) = {π₯ β 2, π₯ < 0 and 1, π₯ β₯ 0, find the value of β«_(β2)^(3) π(π₯) dπ₯.

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### Video Transcript

For the piecewise function π of π₯ equals π₯ minus two when π₯ is less than zero and one when π₯ is greater than or equal to zero, find the value of the definite integral between negative two and three of π of π₯ with respect to π₯.

Weβre working with a piecewise function, so weβre going to need to be a little bit careful when integrating our function with respect to π₯. Now, in fact, what we do is separate our integral into two separate integrals based on the boundaries of our original integral. Our definite integral is from negative two to three. But from negative two to zero, the function is defined as π₯ minus two. But between zero and three, the function is defined as one. And so, the definite integral between negative two and three of π of π₯ with respect to π₯ is equal to the definite integral between negative two and zero of π₯ minus two with respect to π₯ plus the definite integral between zero and three of one with respect to π₯.

Letβs begin by finding our first integral. Remember, when integrating a power term whose exponent is not equal to negative one, we add one to the exponent and then divide by that new value. So, the integral of π₯ is π₯ squared over two. And the integral of negative two is negative two π₯. The integral of one is simply π₯. So, the definite integral between negative two and three of π of π₯ with respect to π₯ is π₯ squared over two minus two π₯ evaluated between negative two and zero and π₯ evaluated between zero and three. So, thatβs zero squared over two minus two times zero minus negative two squared over two minus two times negative two plus three times zero. Zero squared over two is zero. And two times zero is zero.

So, this simplifies to negative two plus four plus three, which is negative six plus three, which is negative three. And so, the definite integral between negative two and three of π of π₯ with respect to π₯ is negative three.