The Fundamental Theorem of Calculus, Part 2<\/h3>\n
If [latex]f[\/latex] is continuous over the interval [latex]\\left[a,b\\right][\/latex] and [latex]F(x)[\/latex] is any antiderivative of [latex]f(x),[\/latex] then<\/p>\n
We often see the notation [latex]{F(x)|}_{a}^{b}[\/latex] to denote the expression [latex]F(b)-F(a).[\/latex]<\/p>\n
We use this vertical bar and associated limits [latex]a[\/latex] and [latex]b[\/latex] to indicate that we should evaluate the function [latex]F(x)[\/latex] at the upper limit (in this case, [latex]b[\/latex]), and subtract the value of the function [latex]F(x)[\/latex] evaluated at the lower limit (in this case, [latex]a[\/latex]).<\/p>\n<\/section>\n
The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting.<\/p>\n Proof<\/strong><\/p>\n Let [latex]P=\\left\\{{x}_{i}\\right\\},i=0,1\\text{,\u2026,}n[\/latex] be a regular partition of [latex]\\left[a,b\\right].[\/latex] Then, we can write<\/p>\n Now, we know F<\/em> is an antiderivative of [latex]f[\/latex] over [latex]\\left[a,b\\right],[\/latex] so by the Mean Value Theorem for [latex]i=0,1\\text{,\u2026,}n[\/latex] we can find [latex]{c}_{i}[\/latex] in [latex]\\left[{x}_{i-1},{x}_{i}\\right][\/latex] such that<\/p>\n Then, substituting into the previous equation, we have<\/p>\n Taking the limit of both sides as [latex]n\\to \\infty ,[\/latex] we obtain<\/p>\n [latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n Use the second part of the Fundamental Theorem of Calculus to evaluate<\/p>\n [reveal-answer q=\"fs-id1170572173704\"]Show Solution[\/reveal-answer] Recall the power rule for antiderivatives:<\/p>\n Use this rule to find the antiderivative of the function and then apply the theorem. We have<\/p>\n Analysis<\/strong><\/p>\n Notice that we did not include the \u201c+ C<\/em>\u201d term when we wrote the antiderivative. The reason is that, according to the Fundamental Theorem of Calculus, Part 2, any<\/em> antiderivative works. So, for convenience, we chose the antiderivative with [latex]C=0.[\/latex] If we had chosen another antiderivative, the constant term would have canceled out. This always happens when evaluating a definite integral.<\/p>\n The region of the area we just calculated is depicted in Figure 3. Note that the region between the curve and the [latex]x[\/latex]-axis is all below the [latex]x[\/latex]-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.<\/p>\n\n\n[caption id=\"\" align=\"aligncenter\" width=\"325\"] Use the second part of the Fundamental Theorem of Calculus to evaluate [latex]{\\displaystyle\\int }_{1}^{2}{x}^{-4}dx.[\/latex]<\/p>\n [reveal-answer q=\"fs-id1170571639103\"]Show Solution[\/reveal-answer] [latex]\\frac{7}{24}[\/latex]<\/p>\n [\/hidden-answer]<\/p>\n<\/section>\n James and Kathy are racing on roller skates. They race along a long, straight track, and whoever has gone the farthest after [latex]5[\/latex] sec wins a prize. If James can skate at a velocity of [latex]f(t)=5+2t[\/latex] ft\/sec and Kathy can skate at a velocity of [latex]g(t)=10+ \\cos (\\frac{\\pi }{2}t)[\/latex] ft\/sec, who is going to win the race?<\/p>\n\n[reveal-answer q=\"fs-id1170572332613\"]Show Solution[\/reveal-answer] We need to integrate both functions over the interval [latex]\\left[0,5\\right][\/latex] and see which value is bigger. For James, we want to calculate<\/p>\n Using the power rule, we have<\/p>\n Thus, James has skated [latex]50[\/latex] ft after [latex]5[\/latex] sec. Turning now to Kathy, we want to calculate<\/p>\n We know [latex] \\sin t[\/latex] is an antiderivative of [latex] \\cos t,[\/latex] so it is reasonable to expect that an antiderivative of [latex] \\cos (\\frac{\\pi }{2}t)[\/latex] would involve [latex] \\sin (\\frac{\\pi }{2}t).[\/latex] However, when we differentiate [latex] \\sin (\\frac{\\pi }{2}t),[\/latex] we get [latex]\\frac{\\pi }{2} \\cos (\\frac{\\pi }{2}t)[\/latex] as a result of the chain rule, so we have to account for this additional coefficient when we integrate. We obtain<\/p>\n Kathy has skated approximately [latex]50.6[\/latex] ft after [latex]5[\/latex] sec. Kathy wins, but not by much!<\/p>\n Watch the following video to see the worked solution to this example.<\/p>\n
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\n[hidden-answer a=\"fs-id1170572173704\"]<\/p>\n![\"The](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204110\/CNX_Calc_Figure_05_03_004.jpg\")
Figure 3. The evaluation of a definite integral can produce a negative value, even though area is always positive.[\/caption]\n[\/hidden-answer]\n<\/section>\n
\n[hidden-answer a=\"fs-id1170571639103\"]<\/p>\n
\n[hidden-answer a=\"fs-id1170572332613\"]\n\n\n