Corrected exercises: the Fundamental Theorem of Calculus and net change (MATH 141)

By additivity of the integral, $g(x + h) - g(x) = \int_a^{x+h} f(t)\,dt - \int_a^x f(t)\,dt = \int_x^{x+h} f(t)\,dt$: exactly the thin strip of the figure. On $[x, x + h]$ the continuous function $f$ reaches a minimum $m_h$ and a maximum $M_h$, so the strip lies between two rectangles: $m_h h \le g(x + h) - g(x) \le M_h h$ for $h > 0$. Dividing by $h$ gives $m_h \le \frac{g(x + h) - g(x)}{h} \le M_h$. As $h \to 0$ the interval shrinks to the point $x$, and by continuity both $m_h$ and $M_h$ tend to $f(x)$: the squeeze theorem gives $g'(x) = f(x)$ (the case $h < 0$ is the same with the inequalities reversed). In words: at $x$, the area grows at the rate given by the HEIGHT of the curve at $x$. This is the one place where the continuity of $f$ is used, and it is why FTC 1 requires it.

$f(t) = \sqrt{1 + t^3}$ is continuous for $t \ge -1$, so on every interval $[0, x]$ with $x \ge 0$, and FTC 1 applies directly: $g'(x) = \sqrt{1 + x^3}$. Then $g'(2) = \sqrt{1 + 8} = 3$. Looking for an antiderivative first is a waste twice over: FTC 1 never needs one, since the answer is read off the integrand, and $\sqrt{1 + t^3}$ has no antiderivative expressible with elementary functions anyway. An integrand chosen to be non-integrable is a signal: the question is about FTC 1, not about FTC 2. The classic slip is to write $\sqrt{1 + x^3} - \sqrt{1 + 0^3} = \sqrt{1 + x^3} - 1$, subtracting the value at the lower bound as if it were an evaluation. A CONSTANT lower bound contributes nothing to the derivative.

FTC 1 is stated with the variable in the UPPER bound, so first swap the bounds: $y = \int_x^{\pi} \cos(t^2)\,dt = -\int_{\pi}^{x} \cos(t^2)\,dt$. The integrand is continuous everywhere, so $\frac{dy}{dx} = -\cos(x^2)$. Sanity check with the picture: moving the lower bound $x$ to the right REMOVES area where $\cos(t^2) > 0$, so $y$ should decrease there, and the minus sign says exactly that. Two wrong answers are common: $\cos(x^2)$, the sign lost, and $\cos(\pi^2) - \cos(x^2)$, an evaluation instead of a derivative; $\cos(\pi^2)$ is a constant with nothing to do with the rate.

Here the upper bound is a function of $x$. Write $y = G(u)$ with $G(u) = \int_1^{u} e^{t^2}\,dt$ and $u = \sqrt{x}$. By FTC 1, $G'(u) = e^{u^2}$, and by the chain rule $\frac{dy}{dx} = G'(u)\,\frac{du}{dx} = e^{(\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{x}}{2\sqrt{x}}$. The integrand is evaluated AT THE BOUND $\sqrt{x}$, not at $x$: $e^{x^2}$ would be wrong. And the factor $\frac{1}{2\sqrt{x}}$, the derivative of the bound, is owed every time the bound is not simply $x$; forgetting it is the most frequent loss on this question, usually half the marks. Note that $e^{t^2}$ has no elementary antiderivative either.

For $x \neq 0$ we have $x^2 > 0$, so $\frac{1}{t}$ is continuous on the interval between $1$ and $x^2$, and $\ln t$ is an antiderivative there: $F(x) = \Big[\ln t\Big]_1^{x^2} = \ln(x^2) - \ln 1 = \ln(x^2) = 2\ln|x|$. The absolute value is not decoration: for $x = -3$, $\ln(x^2) = \ln 9 = 2\ln 3$, while $2\ln x$ is not even defined. Differentiating, $\frac{d}{dx}\,2\ln|x| = \frac{2}{x}$ for every $x \neq 0$. By FTC 1 and the chain rule, $F'(x) = \frac{1}{x^2} \cdot 2x = \frac{2}{x}$. The two routes agree, as they must: FTC 1 is the shortcut that skips the antiderivative, and here it can be checked because the antiderivative happens to be known.

The integrand $f(t) = \sqrt{1 + t^4}$ is continuous everywhere, so we may split at $0$: $h(x) = \int_0^{x^2} f(t)\,dt - \int_0^{x} f(t)\,dt$. FTC 1 with the chain rule on the first piece, FTC 1 alone on the second: $h'(x) = f(x^2) \cdot 2x - f(x) \cdot 1 = 2x\sqrt{1 + x^8} - \sqrt{1 + x^4}$. Note $f(x^2) = \sqrt{1 + (x^2)^4} = \sqrt{1 + x^8}$: the bound replaces $t$ everywhere, including inside the power. At $x = 1$: $h'(1) = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$. The general rule, worth knowing in this form: $\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)\,dt = f(v(x))\,v'(x) - f(u(x))\,u'(x)$.

The student evaluated the integrand at both bounds but did not multiply by the derivatives of the bounds. The factor for the lower bound is $1$, so nothing is lost there; the factor $2x$ for the upper bound is missing, and it spoils the answer everywhere except at $x = \frac{1}{2}$, where $2x = 1$ by accident. At $x = 1$ both bounds equal $1$, so $h(1) = \int_1^1 f = 0$. The student's value $0$ would say that $h$ has a horizontal tangent there; the true slope is $h'(1) = \sqrt{2}$, and the tangent line is $y = \sqrt{2}\,(x - 1)$. The picture confirms it: for $x$ slightly larger than $1$, $x^2 > x$ and $h(x) > 0$; for $x$ slightly smaller, $x^2 < x$, the bounds are in decreasing order and $h(x) < 0$. So $h$ crosses zero at $x = 1$ with a positive slope.

The integrand $f(t) = \frac{1}{\sqrt{1 - t^2}}$ is continuous on $(-1, 1)$, and for $0 < x < \frac{\pi}{2}$ both bounds $\sin x$ and $\cos x$ lie in $(0, 1)$, so the whole interval between them stays inside $(-1, 1)$. By the rule of a): $k'(x) = f(\sin x)\cos x - f(\cos x)(-\sin x) = \frac{\cos x}{\sqrt{1 - \sin^2 x}} + \frac{\sin x}{\sqrt{1 - \cos^2 x}}$. Now $\sqrt{1 - \sin^2 x} = \sqrt{\cos^2 x} = |\cos x|$, NOT $\cos x$ in general; here $\cos x > 0$ on $(0, \frac{\pi}{2})$, so it equals $\cos x$. Likewise $\sqrt{1 - \cos^2 x} = \sin x$. Hence $k'(x) = 1 + 1 = 2$. With FTC 2: $\arcsin$ is an antiderivative of $f$ on $(-1, 1)$, so $k(x) = \arcsin(\sin x) - \arcsin(\cos x)$. Since $x \in (0, \frac{\pi}{2})$, $\arcsin(\sin x) = x$; and $\cos x = \sin(\frac{\pi}{2} - x)$ with $\frac{\pi}{2} - x \in (0, \frac{\pi}{2})$, so $\arcsin(\cos x) = \frac{\pi}{2} - x$. Therefore $k(x) = 2x - \frac{\pi}{2}$, whose derivative is indeed $2$. Both simplifications, $\sqrt{\cos^2 x} = \cos x$ and $\arcsin(\sin x) = x$, hold ONLY on the stated interval: that is why the statement gives it.

$\frac{1}{t}$ is continuous on $(0, \infty)$, and for $x > 0$ both $x$ and $ax$ are positive. Split at $1$: $H(x) = \int_1^{ax} \frac{dt}{t} - \int_1^{x} \frac{dt}{t} = L(ax) - L(x)$. By FTC 1 and the chain rule, $H'(x) = \frac{1}{ax} \cdot a - \frac{1}{x} = \frac{1}{x} - \frac{1}{x} = 0$ for every $x > 0$. A function with zero derivative on an INTERVAL is constant there (mean value theorem, MATH 140), and $(0, \infty)$ is an interval, so $H(x) = H(1) = \int_1^{a} \frac{dt}{t} = L(a)$. Comparing the two expressions of $H(x)$: $L(ax) - L(x) = L(a)$, that is $L(ax) = L(a) + L(x)$. The product law of logarithms has just been proved from the integral alone. The word interval matters: the same argument on a domain in two pieces would only give one constant on each piece.

Take $a = \frac{1}{x}$ in d): $L(1) = L(\frac{1}{x}) + L(x)$, and $L(1) = \int_1^1 \frac{dt}{t} = 0$, so $L(\frac{1}{x}) = -L(x)$. Next, on $[1, 2]$ we have $t \le 2$, so $\frac{1}{t} \ge \frac{1}{2}$, and the comparison property of the integral gives $L(2) = \int_1^2 \frac{dt}{t} \ge \frac{1}{2}(2 - 1) = \frac{1}{2}$ (indeed $\ln 2 \approx 0.69$). Applying d) with $a = 2$ repeatedly, $L(2^n) = L(2) + L(2^{n-1}) = \dots = nL(2) \ge \frac{n}{2}$. Finally $L'(x) = \frac{1}{x} > 0$ by FTC 1, so $L$ is increasing, and $x \ge 2^n$ forces $L(x) \ge \frac{n}{2}$. Since $n$ is arbitrary, $L(x) \to \infty$ as $x \to \infty$, and by $L(\frac{1}{x}) = -L(x)$, $L(x) \to -\infty$ as $x \to 0^+$. Note that the defining integral is NEGATIVE for $0 < x < 1$: the bounds are then in decreasing order, even though the integrand is positive.

Each value is a signed area, computed by geometry since $f$ is made of familiar shapes. $g(0) = \int_0^0 f = 0$. On $[0, 2]$ the region is a quarter of a disc of radius $2$: $g(2) = \frac{1}{4}\pi (2)^2 = \pi$. On $[0, 4]$, the half disc: $g(4) = \frac{1}{2}\pi(2)^2 = 2\pi$. On $[4, 6]$ the graph is BELOW the axis, forming a triangle of base $2$ and height $2$, area $2$, counted negatively: $g(6) = 2\pi - 2$. On $[6, 8]$ another such triangle: $g(8) = 2\pi - 4$. With $\pi \approx 3.14$, $g(8) \approx 2.28 > 0$. The sign convention is the whole point: $g$ is a NET accumulation, and the parts below the axis subtract.

$f$ is continuous on $[0, 8]$, so by FTC 1, $g'(x) = f(x)$. On $(0, 4)$, $f > 0$: $g$ is increasing on $[0, 4]$. On $(4, 8)$, $f < 0$: $g$ is decreasing on $[4, 8]$. So $g$ has its absolute maximum where $f$ changes sign from positive to negative, at $x = 4$, with $g(4) = 2\pi$. Since $g$ rises then falls, the absolute minimum is at an endpoint: compare $g(0) = 0$ and $g(8) = 2\pi - 4$. Because $\pi > 2$, $2\pi - 4 > 0$, so the minimum is $0$, at $x = 0$. Writing the comparison of the two endpoints is what the marker looks for; placing the minimum at $x = 8$ because $g$ decreases there is the typical error.

Where $f$ is differentiable, $g''(x) = f'(x)$. $f$ increases on $(0, 2)$, the left half of the semicircle, and on $(6, 8)$: $g$ is concave up there. $f$ decreases on $(2, 4)$ and on $(4, 6)$: $g$ is concave down on $(2, 6)$. The concavity changes at $x = 2$ and at $x = 6$, so the inflection points are $(2, \pi)$ and $(6, 2\pi - 2)$: they are the points where $g' = f$ has a local maximum and a local minimum. At $x = 4$ the semicircle has a vertical tangent and meets a segment, so $f'(4)$ does not exist and neither does $g''(4)$; but $g$ is concave down on BOTH sides of $4$, so there is no inflection there. What happens at $4$ is the maximum found in b). The figure of the solution draws $g$ with these three points.

$g'(5) = f(5)$. On $[4, 6]$ the segment from $(4, 0)$ to $(6, -2)$ has equation $y = -(t - 4)$, so $f(5) = -1$ and $g'(5) = -1$. Its slope is $-1$, so $g''(5) = f'(5) = -1$: at $x = 5$, $g$ is decreasing and concave down. Finally $g'(2) = f(2) = 2$, the top of the semicircle: the largest slope $g$ ever has.

The largest value of $f$ is the largest SLOPE of $g$, not its largest value. At $x = 2$, $g$ is climbing as fast as it ever does; it keeps climbing until $x = 4$, since $f$ stays positive on $(2, 4)$, so $g(4) = 2\pi > g(2) = \pi$ and $x = 2$ cannot be a maximum. The classmate confused $f$ with $g$: the maximum of an accumulation is where the RATE changes sign from positive to negative, while the maximum of the rate is an inflection point of the accumulation. The tangent line at $x = 2$ has slope $g'(2) = 2$ and passes through $(2, \pi)$: $y = \pi + 2(x - 2)$, that is $y = 2x + \pi - 4$.

$\sqrt{x} > 0$ on $[1, 4]$, so the integrand is continuous there. Divide term by term: $\frac{x^2 + 1}{\sqrt{x}} = x^{3/2} + x^{-1/2}$. The power rule gives $F(x) = \frac{2}{5}x^{5/2} + 2x^{1/2}$, and $F'(x) = x^{3/2} + x^{-1/2}$ checks it. Then $F(4) = \frac{2}{5}(32) + 2(2) = \frac{84}{5}$, using $4^{5/2} = (\sqrt{4})^5 = 32$, and $F(1) = \frac{2}{5} + 2 = \frac{12}{5}$. Answer: $\frac{84}{5} - \frac{12}{5} = \frac{72}{5}$. The error to avoid is integrating the numerator and the denominator separately, $\frac{x^3/3 + x}{\frac{2}{3}x^{3/2}}$: there is no quotient rule for integrals, and differentiating that expression does not give back the integrand.

On $[0, \frac{\pi}{4}]$, $\cos\theta \ge \frac{\sqrt{2}}{2} > 0$, so the integrand is continuous. Split the fraction: $\frac{1 + \cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta} + 1 = \sec^2\theta + 1$. An antiderivative is $\tan\theta + \theta$. So the integral is $\left(\tan\frac{\pi}{4} + \frac{\pi}{4}\right) - (\tan 0 + 0) = 1 + \frac{\pi}{4}$. Sanity check: the integrand is between $2$ (at $\theta = 0$) and $3$ (at $\frac{\pi}{4}$), over an interval of length $\frac{\pi}{4} \approx 0.79$, so the value lies between $1.57$ and $2.36$; and $1 + \frac{\pi}{4} \approx 1.79$.

On $[0, \frac{1}{2}]$, $1 - x^2 \ge \frac{3}{4} > 0$, so the integrand is continuous. It is a table entry times a constant: an antiderivative is $4\arcsin x$. The value is $4\left(\arcsin\frac{1}{2} - \arcsin 0\right) = 4 \cdot \frac{\pi}{6} = \frac{2\pi}{3}$. The trap is the exact value: $\arcsin\frac{1}{2} = \frac{\pi}{6}$, because $\sin\frac{\pi}{6} = \frac{1}{2}$; students who write $\frac{\pi}{3}$ are reading the cosine table. Bound check: the integrand runs from $4$ to $\frac{4}{\sqrt{3/4}} = \frac{8}{\sqrt{3}} \approx 4.62$ over a length $\frac{1}{2}$, so the value is between $2$ and $2.31$; $\frac{2\pi}{3} \approx 2.09$ fits, while $\frac{4\pi}{3} \approx 4.19$ would not.

$x > 0$ on $[1, 2]$: continuous. Expand, then divide: $\frac{(x - 1)^2}{x} = \frac{x^2 - 2x + 1}{x} = x - 2 + \frac{1}{x}$. An antiderivative is $\frac{x^2}{2} - 2x + \ln x$, with no absolute value needed since $x > 0$ here. Evaluate: $(2 - 4 + \ln 2) - (\frac{1}{2} - 2 + 0) = \ln 2 - \frac{1}{2}$. Sign check: the integrand is $\ge 0$, so the answer must be positive, and with $\ln 2 \approx 0.69$ it is about $0.19$. It must also be less than $\frac{1}{2}$, since the integrand is at most $\frac{1}{2}$ on $[1, 2]$, its value at $x = 2$. The pitfall is to see $(x - 1)^2$ and reach for a substitution; the question is solved by one line of algebra.