M408C: The Indefinite Integral and the NetChange TheoremOctober 28, 2008The Fundamental Theorem of Calculus allows us to think of the antiderivative of a functionf(x) as the integral of f. This motivates the definition of the indefinite integral of f:Note that the antiderivative we choose will always have a constant, so for example:Rx2dx =x3/3 + C. Note als o that the definite integral of a function is a number whereas the indefiniteintegral of a function is a function. The connection between them is given by FTC2:We can rewrite FTC2 using the fact that F0= f and get the following theorem.Theorem 1 (The Net Change Theorem). The integral of a rate of chan ge is the net change:1.2.13. (5.3.32) EvaluateRπ/40sec θ tan θ dθ.Solution:Zπ/40sec θ tan θ dθ = sec θπ/40= secπ4− sec 0 =1cosπ4−1cos 0=11√2− 1 =√2 − 1.4. (5.4.38) Find the indefinite integralR(1 + x2)3dx and use it to evaluate the definite integralR10(1 + x2)3dx.Solution: We can try gues sing the antiderivative of f(x) = (1 + x2)3is F (x) =14(1 + x2)4+ C,but differentiating F shows quickly that is not the case. Instead we do some algebra:Z(1 + x2)3dx =Z(1 + 2x2+ x4)(1 + x2) dx =Z(1 + 3x2+ 3x4+ x6) dx=Z1 dx +Z3x2dx +Z3x4dx +Zx6dx= x + x3+35x5+17x7+ C = F (x)To determine the definite integral, we do the following:Z10(1 + x2)3dx =Z(1 + x2)3dx10= F (1) − F (0)= 1 + 13+3513+1717+ C −0 + 03+3505=1707+ C= 2 +35+17=9635.5. (5.4.58). Suppose the acceleration of a particle is given by a(t) = 2t + 3 and its initial velocityis v(0) = −4. Find the velocity of the particle at time t and determine the total distance traveledby the object for 0 ≤ t ≤ 3.Solution: Note first that we want the total distance, not the net distance. That is to say,we don’t want the “negative” area under our curve to cancel with the “positive” area. In otherwords, we want the “negative” area to be considered “positive” – we do this with the absolutevalue function. Hence, given our velocity function v(t) = s0(t), we want to determine the integralR30|v(t)|dt. First let’s determine what v(t) is:v(t) =Za(t) dt =Z(2t + 3) dt = t2+ 3t + C.Since v(0) = −4, we know that C = −4. Thus s0(t) = v(t) = t2+ 3t −4. Observe that for t ∈ [0, 1],v(t) ≤ 0 and for t ∈ [1, 3], v(t) ≥ 0. Thus our integral becomesZ30|v(t)|dt =Z10−v(t) dt +Z31v(t) dt.Plugging in for v(t), we haveZ30|v(t)|dt =Z10−(t2+ 3t − 4) dt +Z31(t2+ 3t − 4) dt = −13t3+32t2− 4t10+13t3+32t2− 4t31=−13−32+ 4+273+272− 12 −13−32+ 4=−2 − 9 + 24 + 54 + 81 − 72 − 2 − 9 +
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