4.1: Antidifferentiation

1. This problem asks: "The derivative of what function is \(\cos x \ques\)" One such function is \(\sin x.\) But the derivative of \(\sin x\) plus any constant—for example, the derivative of \((\sin x + 1)\) or of \((\sin x - 100)\)—is also \(\cos x.\) Thus, if \(C\) is any constant, then the general antiderivative of \(\cos x\) is \(\sin x + C.\) So \[\int \cos x \di x = \boxed{\sin x + C}\]

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2. The derivative of \(e^x\) is \(e^x,\) so the converse statement is that \(e^x\) is an antiderivative of \(e^x.\) We always add a constant of integration, so \[\int e^x \di x = \boxed{e^x + C}\]

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3. Since the derivative of \(\ln|x|\) is \(1/x,\) an antiderivative of \(1/x\) is \(\ln|x|.\) We then add an arbitrary constant in the final answer: \[\int \frac{1}{x} \di x = \boxed{\ln|x| + C}\]

1. We use the Reverse Power Rule, following \(\eqref{eq:reverse-power-rule},\) to obtain \[\int x^{-2} \di x = \frac{x^{-2 + 1}}{-2 + 1} + C = \boxed{-\frac{1}{x} + C}\]

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2. We apply a combination of rules, as follows: \[ \baat{2} \int \par{5x^4 + \cos x} \di x &= \int 5x^4 \di x + \int \cos x \di x \comment{\text{by Sum Rule}} \nl &= 5 \int x^4 \di x + \int \cos x \di x \comment{\text{by Constant Multiple Rule}} \nl &= 5 \par{\frac{x^{4 + 1}}{4 + 1}} + \int \cos x \di x \comment{\text{by Reverse Power Rule}} \nl &= \boxed{x^5 + \sin x + C} \eaat \]

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3. We first write \[ \ba \int \par{\sqrt{t} - 2 \sin t + \frac{4}{1 + t^2}} \di t &= \int \par{t^{1/2} - 2 \sin t + \frac{4}{1 + t^2}} \di t \nl &= \int t^{1/2} \di t - 2 \int \sin t \di t + 4 \int \frac{1}{1 + t^2} \di t \cma \ea \] where the last step is true by the Sum and Difference Rules combined with the Constant Multiple Rule. Using the Reverse Power Rule to antidifferentiate \(t^{1/2}\) then gives \[\frac{t^{3/2}}{3/2} + 2 \cos t + 4 \atan t + C = \boxed{\tfrac{2}{3} t^{3/2} + 2 \cos t + 4 \atan t + C}\]

Antidifferentiating \(f'\) gives \begin{equation*} f(x) = e^x + x^2 + 7x + C \cma \end{equation*} which is the general solution to the differential equation. To find the particular solution, we substitute the given initial condition, \(f(0) = 4,\) to obtain \[ \ba f(0) = e^0 + (0)^2 + 7(0) + C &= 4 \nl \implies C &= 3 \pd \ea \] Thus, the particular solution satisfying \(f(0) = 4\) is \begin{equation} f(x) = \boxed{e^x + x^2 + 7x + 3} \pd \label{eq:ex-diff-eq-f(x)-particular} \end{equation}

Verification Checking our answer is simple: It's easy to see that \(\eqref{eq:ex-diff-eq-f(x)-particular}\) satisfies \(f(0) = 4 \col\) \[f(0) = e^0 + 0^2 + 7(0) + 3 \equalsCheck 4 \pd\] Moreover, differentiating \(\eqref{eq:ex-diff-eq-f(x)-particular}\) yields \[f'(x) = e^x + 2x + 7 \cma\] which matches the given differential equation.

• When \(0 \lt x \lt 1\) and \(x \gt 3,\) \(f\) is negative, meaning the slope of \(F\) is negative. Thus, \(F\) is decreasing on these intervals.
• For \(1 \lt x \lt 3,\) \(f\) is positive, so the slope of \(F\) is positive. Therefore, \(F\) is increasing on this interval.
• Since \(f(1) = 0\) and \(f(3) = 0,\) \(F\) has horizontal tangents when \(x = 1\) and \(x = 3.\) By the First Derivative Test (see Section 3.1), \(F(1)\) is a relative minimum because \(F' = f\) changes sign from negative to positive at \(x = 1.\) Similarly, \(F(3)\) is a relative maximum because \(F' = f\) changes sign from positive to negative at \(x = 3.\)

• For \(0 \lt x \lt 2,\) \(F'' = f' \gt 0\) (since \(f\) is increasing), so \(F\) is concave up on this interval. In contrast, \(F'' = f' \lt 0\) when \(x \gt 2\) (since \(f\) is decreasing), so \(F\) is concave down for \(x \gt 2.\)
• \(F'' = f'\) switches sign at \(2,\) so \(F\) has an inflection point when \(x = 2.\)

Velocity Function Antidifferentiating \(a(t) = 4t + 6\) yields \[ \ba v(t) &= \int (4t + 6) \di t \nl &= 2t^2 + 6t + C \cma \ea \] where \(C\) is a constant. The particle's initial velocity is \(v(0) = 3,\) which we substitute to get \[ \ba v(0) = 2(0)^2 + 6(0) + C &= 3 \nl \implies C &= 3 \pd \ea \] Thus, the velocity function is \[ v(t) = 2t^2 + 6t + 3 \pd \]

Position Function Antidifferentiating the velocity function shows \[ \ba x(t) &= \int \par{2t^2 + 6t + 3} \di t \nl &= \tfrac{2}{3} t^3 + 3t^2 + 3t + D \cma \ea \] where \(D\) is a constant. The initial position is \(x(0) = 2,\) which we substitute to attain \[ \ba x(0) = \tfrac{2}{3}(0)^3 + 3(0)^2 + 3(0) + D &= 2 \nl \implies D &= 2 \pd \ea \] Thus, the position function is \[ x(t) = \boxed{\tfrac{2}{3} t^3 + 3t^2 + 3t + 2} \] One may check this answer by noting that \[\andThree{x''(t) = a(t) \equalsCheck 4t + 6}{x(0) \equalsCheck 2}{x'(0) = v(0) \equalsCheck 3} \pd\]

Since the rock is dropped, its initial velocity is \(v(0) = 0 \undiv{ft}{sec}.\) We take the positive direction to be upward; the rock's velocity and acceleration are therefore both negative as it falls. Also, the object's initial position—its height above the ground, \(h(t)\)—is \(h(0) = 100 \un{ft}.\)

Velocity Function The acceleration due to gravity is a constant \(-32 \undiv{ft}{sec}^2,\) so the velocity function is \[ \ba v(t) &= \int (-32) \di t \nl &= -32t + C \cma \ea \] where \(C\) is a constant. Substituting the initial condition \(v(0) = 0 \undiv{ft}{sec}\) shows \[ \ba v(0) = -32(0) + C &= 0 \nl \implies C &= 0 \pd \ea \] Thus, the rock's velocity function is \[v(t) = -32t \pd\]

Height Function The object's position function (its height above the ground) is given by antidifferentiating \(v(t) \col\) \[ \ba h(t) &= \int (-32 t) \di t \nl &= -16t^2 + D \ea \] for some constant \(D.\) Substituting the initial condition \(h(0) = 100 \un{ft},\) we see \[ \ba h(0) = -16(0)^2 + D &= 100 \nl \implies D &= 100 \pd \ea \] Thus, the height function is \[ h(t) = \boxed{-16t^2 + 100} \]

Immediately Before Impact When the rock reaches the ground, \(h(t) = 0 \un{ft}.\) Solving for the positive value of \(t,\) we have \[ \ba h(t) = -16t^2 + 100 &= 0 \nl \implies t &= 2.5 \un{sec} \pd \ea \] The rock therefore hits the ground after \(2.5 \text{sec}.\) The velocity at this time is \[v(2.5) = -32(2.5) = -80 \undiv{ft}{sec} \pd\] Speed is the magnitude (absolute value) of velocity, so the rock's final speed is \[\abs{v(2.5)} = \abs{-80} = \boxed{80 \undiv{ft}{sec}}\]

1. If the positive direction is upward, then the ball's initial velocity is \(v(0) = 16 \undiv{ft}{sec},\) and the acceleration due to gravity is \(-32 \undiv{ft}{sec}^2.\) Thus, the velocity function is \[ \ba v(t) &= \int (-32) \di t \nl &= -32 t + C \pd \ea \] Substituting the initial condition \(v(0) = 16 \undiv{ft}{sec}\) gives \[ \ba v(0) = -32(0) + C &= 16 \nl \implies C &= 16 \pd \ea \] Therefore, \[ v(t) = \boxed{-32t + 16} \]

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2. The ball reaches its maximum height when it is momentarily stationary—that is, when \(v(t) = 0 \undiv{ft}{sec} \col\) \[ \ba -32t + 16 &= 0 \nl \implies t &= \boxed{0.5 \un{sec}} \ea \]

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3. The position function—the height above the ground, \(h(t)\)—is given by integrating \(v(t) \col\) \[ \ba h(t) &= \int \par{-32t + 16} \di t \nl &= -16t^2 + 16t + D \ea \] for some constant \(D.\) The ball is initially at the ground, so \(h(0) = 0 \un{ft}.\) Substituting this initial condition gives \[ \ba h(0) = -16(0)^2 + 16(0) + D &= 0 \nl \implies D &= 0 \pd \ea \] So the height as a function of time is \[h(t) = -16t^2 + 16t \pd\] The maximum height reached is when \(t = 0.5 \un{sec} \col\) \[h(0.5) = -16 \par{0.5}^2 + 16 (0.5) = \boxed{4 \un{ft}}\]