Chapter 5 | Integration
557
2 ⎛
3 3 −4
⎞ ⎠ dt = t
∫
⎝ t 2 −4
t | −2 2
−2
⎡ ⎣ (2) 3
⎤ ⎦ −
⎡ ⎣ (−2) 3
⎤ ⎦
(2)
(−2)
=
3 −4
3 −4
⎛ ⎝ 8
⎞ ⎠ −
⎛ ⎝ − 8 3 +8 ⎞ ⎠
=
3 −8
8 3 −8
= 8 3 −8+ = 16 3 −16 = − 32 3 .
Analysis Notice that we did not include the “+ C ” term when we wrote the antiderivative. The reason is that, according to the Fundamental Theorem of Calculus, Part 2, any antiderivative works. So, for convenience, we chose the antiderivative with C =0. If we had chosen another antiderivative, the constant term would have canceled out. This always happens when evaluating a definite integral. The region of the area we just calculated is depicted in Figure 5.28 . Note that the region between the curve and the x -axis is all below the x -axis. Area is always positive, but a definite integral can still produce a negative number (a net signed area). For example, if this were a profit function, a negative number indicates the company is operating at a loss over the given interval.
Figure 5.28 The evaluation of a definite integral can produce a negative value, even though area is always positive.
Example 5.21 Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2
Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2: ∫ 1 9 x −1 x dx . Solution
Made with FlippingBook - professional solution for displaying marketing and sales documents online