Solutions

• Let $F(x) = \int_0^x e^{-t^2} dt$.
  Find $F'(x)$ and $F''(x)$.
  Show that $F$ is an increasing function of $x$.
  Show that $F$ has one point of inflection.
  
  By $FTC$, we have:
  $$F(x) = \int_0^x e^{-t^2} dt,$$
  $$F'(x) = e^{-x^2},$$
  $$F''(x) = \frac{d}{dx} e^{-x^2} = - 2xe^{-x^2}.$$
  $F'$ is always positive, so the function $F$ is increasing.
  
  We have:
  • $F''(x) < 0$, for $x < 0$, so the graph of $F$ is concave down for $x < 0$.
  • $F''(0) = 0$,
  • $F''(x) > 0$, for $x > 0$, so the graph of $F$ is concave up for $x > 0$.
    
  So the graph of $F(x)$ changes concavity, going through $x = 0$.
  So $x = 0$ is a point of inflection.
• Let $G(x) = \int_0^x te^{-t^2} dt$.
  Find the critical points, local maxima and minima and points of inflection of the graph of $G(x)$.
  
  We have, using $FTC$ and the product rule:
  $$G'(x) = xe^{-x^2},$$
  $$G''(x) = e^{-x^2}(1 - 2x^2).$$
  The only critical point is $x = 0$.
  Since $G''(0) = 1 > 0$, the graph of $G$ is concave up at $x = 0$, so $x = 0$ is a local minimum.
  We have:
  • $G''(x) > 0$, for $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$, so the graph of $G$ is concave up for $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$.
  • $G''(\pm \frac{1}{\sqrt{2}}) = 0$,
  • $G''(x) < 0$, for $x > \frac{1}{\sqrt{2}}$ and for $x < -\frac{1}{\sqrt{2}}$, so the graph of $G$ is concave down for $x > \frac{1}{\sqrt{2}}$ and for $x < -\frac{1}{\sqrt{2}}$.
    
  So the graph of $G(x)$ changes concavity, going through $x = \pm \frac{1}{\sqrt{2}}$.
  So $x = \pm \frac{1}{\sqrt{2}}$ are the only inflection points.
• Let $H(x) = \int_0^{x^3} \frac{1}{1 + t^3} dt$.
  Find $H'(x)$.
  
  We have $H(x) = G(u(x))$, where $u(x) = x^3$ and $G(u) = \int_0^u \frac{1}{1 + t^3} dt$.
  By $FTC$, we have $G'(u) = \frac{1}{1 + u^3}$.
  By the chain rule, we have:
  $$H'(x) = G'(u(x))u'(x) = 3x^2\frac{1}{1 + (x^3)^3} = \frac{3x^2}{1 + x^9}.$$

• A student writes the following solution to the integral problem:
  $$I = \int_{-2}^{2} \frac{1}{t^2} dt .$$
  By FTC, we have:
  $$I = G(2) - G(-2),$$
  where $G'(t) = \frac{1}{t^2}$.
  Since $\frac{d}{dt} \frac{1}{t} = - \frac{1}{t^2}$, we may take $G(t) = - \frac{1}{t}$.
  Then we have:
  $$G(2) = - \frac{1}{2}, \hspace{10pt} G(-2) = \frac{1}{2},$$
  $$I = G(2) - G(-2) = - \frac{1}{2} - \frac{1}{2} = - 1.$$
  Is something wrong and if so, what?
  
  There must be something wrong, because the integrand is a positive function, so the integral ought to be positive.
  Looking at the integral, we see that the integration region goes through the origin and near the origin the integrand $\frac{1}{t^2}$ gets arbitrarily large.
  Then our usual approach to the integral breaks down and $FTC$ is not valid for the such integrals.
  In fact, we shall see later that the integral does not exist.
• Calculate the following definite integrals:
  • $\int_0^3 (t^2 + t^4 + t^6)dt$.
    
    We first find $G(t)$ such that $G'(t) = t^2 + t^4 + t^6$.
    So $G(t) = \frac{1}{3}t^3 + \frac{1}{5}t^5 + \frac{1}{7}t^7$ will do.
    Then by $FTC$ the integral is $G(3) - G(0) = \frac{1}{3}3^3 + \frac{1}{5}3^5 + \frac{1}{7}3^7 = \frac{12951}{35}$.
  • $\int_0^1 (1 + t)^2 dt$.
    
    We first find $G(t)$ such that $G'(t) = (1 + t)^2 = 1 + 2t + t^2$.
    So $G(t) = t + t^2 + \frac{1}{3}t^3$ will do.
    Then by $FTC$ the integral is $G(1) - G(0) = 1 + 1 + \frac{1}{3} = \frac{7}{3}$.
    
    Alternatively we might note that $G(t) = \frac{(1 + t)^3}{3}$ will do, since by the Chain rule, it has the correct derivative.
    Then by $FTC$ the integral is $G(1) - G(0) = \frac{2^3}{3} - \frac{1}{3} = \frac{7}{3}$.
    
  • $\int_0^1 (1 + t)^3 dt$.
    
    We first find $G(t)$ such that $G'(t) = (1 + t)^3 = 1 + 3t + 3t^2 + t^3$.
    So $G(t) = t + \frac{3t^2}{2} + t^3 + \frac{1}{4}t^4$ will do.
    Then by $FTC$ the integral is $G(1) - G(0) = 1 +\frac{3}{2} + 1 + \frac{1}{4} = \frac{15}{4}$.
    
    Alternatively we might note that $G(t) = \frac{(1 + t)^4}{4}$ will do, since by the Chain rule, it has the correct derivative.
    Then by $FTC$ the integral is $G(1) - G(0) = \frac{2^4}{4} - \frac{1}{4} = \frac{15}{4}$.
  • $\int_0^1 (1 + t^2)^2 dt$.
    
    We first find $G(t)$ such that $G'(t) = (1 + t^2)^2 = 1 + 2t^2 + t^4$.
    So $G(t) = t + \frac{2t^3}{3} + \frac{1}{5}t^5$ will do.
    Then by $FTC$ the integral is $G(1) - G(0) = 1 + \frac{2}{3} + \frac{1}{5} = \frac{28}{15}$.
  • $\int_0^1t(1 + t^2)^3 dt$.
    
    We first find $G(t)$ such that $G'(t) = t(1 + t^2)^3 = t(1 + 3t^2 + 3t^4 + t^6) = t + 3t^3 + 3t^5 + t^7$.
    So $G(t) = \frac{t^2}{2} + \frac{3t^4}{4} + \frac{t^6}{2} + \frac{t^8}{8}$ will do.
    Then by $FTC$ the integral is $G(1) - G(0) = \frac{1}{2} + \frac{3}{4} + \frac{1}{2} + \frac{1}{8} = \frac{15}{8}$.
    
    Alternatively we might note that $G(t) = \frac{(1 + t^2)^4}{8}$ will do, since by the Chain rule, it has the correct derivative.
    Then by $FTC$ the integral is $G(1) - G(0) = \frac{2^4}{8} - \frac{1^4}{8} = \frac{15}{8}$.
  • $\int_1^3 (t - \frac{1}{t})^2 dt$.
    
    We first find $G(t)$ such that $G'(t) = (t - \frac{1}{t})^2 = t^2 - 2 + \frac{1}{t^2}$.
    So $G(t) = \frac{t^3}{3} - 2t - \frac{1}{t}$ will do.
    Then by $FTC$ the integral is:
    $G(3) - G(1) = \frac{3^3}{3} - 2(3) - \frac{1}{3} - (\frac{1}{3} - 2 - 1) = \frac{16}{3}$.
  • $\int_1^3 \vert t^2 - 4\vert dt$.
    
    The function $t^2 - 4$ is negative in the interval $[1,2]$ and is positive in the interval $[2, 3]$.
    So we split the integral into two pieces:
    $$\int_1^3 \vert t^2 - 4\vert dt = A + B,$$
    $$A = \int_1^2 \vert t^2 - 4\vert dt = \int_1^2 -(t^2 - 4)dt = \int_2^1 (t^2 - 4)dt,$$
    $$B = \int_2^3 \vert t^2 - 4\vert dt = \int_2^3 (t^2 - 4)dt.$$
    To do the integrals, we first find $G(t)$ such that $G'(t) = t^2 -4$.
    Then $A = G(1) - G(2)$ and $B = G(3) - G(2)$, so the integral is $A + B = G(3) - 2G(2) + G(1)$.
    We may take $G(t) = \frac{t^3}{3} - 4t$.
    Then $G(3) = -3$, $G(2) = -\frac{16}{3}$ and $G(1) = - \frac{11}{3}$, so we get:
    $$\int_1^3 \vert t^2 - 4\vert dt = G(3) - 2G(2) + G(1) = - 3+ \frac{32}{3} - \frac{11}{3} = 4.$$
    Note that this is quite different from $\int_1^3 (t^2 -4)dt = G(3) - G(1) = -3 + \frac{11}{3} = \frac{2}{3}$.
  • $\int_0^{\frac{\pi}{4}} (\sec^2(x) + \sin(x) - 2\cos(x)) dx$.
    
    We first find $G(x)$ such that $G'(x) = \sec^2(x) + \sin(x) - 2\cos(x)$.
    So $G(x) = \tan(x) - \cos(x) - 2\sin(x)$ will do.
    Then by $FTC$ the integral is $G(\frac{\pi}{4}) - G(0) = 2 - \frac{3}{\sqrt{2}}$.
    
    
  • $\int_4^9 (\sqrt{t} + \frac{1}{\sqrt{t}}) dt$.
    
    We first find $G(t)$ such that $G'(t) = t^{\frac{1}{2}} + t^{-\frac{1}{2}}$.
    So $G(t) = \frac{2}{3}t^{\frac{3}{2}} + 2t^{\frac{1}{2}}$ will do.
    Then by $FTC$ the integral is:
    $$G(9) - G(4) = \frac{2}{3}(9^{\frac{3}{2}}) + 2(9^{\frac{1}{2}}) - (\frac{2}{3}(4^{\frac{3}{2}}) + 2(4^{\frac{1}{2}}))$$
    $$= \frac{2}{3}(27) + 2(3) - (\frac{2}{3}(8) + 2(2))$$
    $$= \frac{44}{3}.$$

  • $\int_1^{27} \frac{x^2 + x + 1}{\sqrt[3]{x}} dx$.
    
    We first find $G(x)$, such that: $G'(x) = \frac{x^2 + x + 1}{\sqrt[3]{x}} = x^{\frac{5}{3}} + x^{\frac{2}{3}} + x^{-\frac{1}{3}}$.
    So $G(x) = \frac{3}{8}x^{\frac{8}{3}} +\frac{3}{5}x^{\frac{5}{3}} + \frac{3}{2}x^{\frac{2}{3}}$ will do.
    Then by $FTC$, the integral is:
    $$G(27) - G(1) = \frac{3}{8}(27)^{\frac{8}{3}} +\frac{3}{5}(27)^{\f... ...}} + \frac{3}{2}(27)^{\frac{2}{3}} - ( \frac{3}{8} + \frac{3}{5}+ \frac{3}{2})$$
    $$= \frac{3^9}{8} +\frac{3^6}{5} + \frac{3^3}{2} - \frac{99}{40} = \frac{9}{40}(5(2187) + 8(81) + 20(3) - 11) = \frac{13086}{5}.$$

  • $\int_0^4 x^2\sqrt{x} dx$.
    We first find $G(x)$, such that: $G'(x) = x^2\sqrt{x} = x^{\frac{5}{2}}$.
    So $G(x) = \frac{2}{7}x^{\frac{7}{2}}$ will do.
    Then by $FTC$, the integral is:
    $$G(4) - G(0) = \frac{2}{7}4^{\frac{7}{2}} = \frac{2^8}{7} = \frac{256}{7}.$$

  • $\int_0^2 (t - 1)(t -2)(t -3) dt$.
• Let $S(x) = \int_0^x \sin(\pi t^2) dt$.
  Find the critical points and local maxima and minima of the function $S(x)$ and give a rough plot of its graph.
  
  By $FTC$, we have $S'(x) = \sin(\pi x^2)$.
  Then by the chain rule, we have: $S''(x) = 2\pi x\cos(\pi x^2)$.
  The critical points occur when $S'(x) = 0$, so when $\sin(\pi x^2) = 0$, so when $\pi x^2 = n\pi$, for some non-negative integer $n$, so when $x^2 = n$, so when $x = \pm \sqrt{n}$.
  When $x = \pm \sqrt{n}$, we have $S'' = \pm 2\pi\sqrt{n} \cos(\pi n) = \pm (-1)^n 2\pi \sqrt{n}$.
  • First consider the critical point $x = 0$.
    We have $S''(0) =0$, so the second derivative test is inapplicable.
    However for small $\vert t\vert$, $\sin(t)$ is approximately $t$, so when $x$ is small and non-zero, $S'(x)$ is approximately $\pi x^2$ so $S'(x)$ is small and positive, for $x\ne 0$ near zero.
    So $S(x)$ is increasing near $x = 0$, so the critical point $x = 0$ is an inflection point.
  • When $n$ is even and positive, $S''$ is positive at the critical point $x = \sqrt{n}$, so this is a local minimum.
  • When $n$ is odd and positive, $S''$ is negative at the critical point $x = \sqrt{n}$, so this is a local maximum.
  • When $n$ is even and positive, $S''$ is negative at the critical point $x = -\sqrt{n}$, so this is a local maximum.
  • When $n$ is odd and positive, $S''$ is positive at the critical point $x = -\sqrt{n}$, so this is a local minimum.
  So we have local maxima at $x = \sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \dots$ and at $x = -1, -\sqrt{3}, -\sqrt{5}, \dots$.
  Also we have local minima at $x = -\sqrt{2}, -2, -\sqrt{6}, -\sqrt{8}, \dots$ and at $x = 1, \sqrt{3}, \sqrt{5}, \dots$.
  
  Finally, making the change from $t$ to $u = t^2$, with $du = 2tdt$, so provided $t$ is positive, $dt = \frac{du}{2t} = \frac{du}{2\sqrt{u}}$ we have, for $x > 0$:
  $$S(x) = \int_0^{x^2}\sin(\pi u)\frac{du}{2\sqrt{u}}.$$
  Then the change in $x$ from the local max at $x = \sqrt{n}$ to the local max at $x = \sqrt{n+2}$ is, for $n = 2k - 1$ and $k$ a positive integer:
  $$\int_n^{n+2}\sin(\pi u)\frac{du}{2\sqrt{u}} = \int_n^{n+1} \sin(\pi u)\frac{du}{2\sqrt{u}} + \int_{n+1}^{n+2} \sin(\pi u)\frac{du}{2\sqrt{u}}$$
  $$= \int_n^{n+1} \sin(\pi u)\frac{du}{2\sqrt{u}} + \int_{n}^{n+1} \sin(\pi (u+1))\frac{du}{2\sqrt{u+1}}$$
  $$= \frac{1}{2}\int_n^{n+1} \sin(\pi u)(\frac{1}{\sqrt{u}} - \frac{1}{\sqrt{u+1}} )du.$$
  $$= - \frac{1}{2}\int_0^{1} \sin(\pi u)(\frac{1}{2\sqrt{u+2k-1}} - \frac{1}{2\sqrt{u+2k}})du.$$
  Here we have used the fact that $\sin(t + \pi) = - \sin(t)$, for any real $t$.
  Since $\frac{1}{\sqrt{t}}$ is decreasing on its domain, the factor $\frac{1}{\sqrt{u+2k-1}} - \frac{1}{\sqrt{u+2k}}$ is positive, as is the term $\sin(\pi u)$, in the interval $(0, 1)$.
  So the difference between adjacent maxima is negative.
  So the local maxima form a decreasing sequence (when $x > 0$).
  Similarly we can show that the local minima, for $x > 0$ form an increasing sequence, each local minimum is below all the local maxima and the gap between the local maxima and minima shrinks to zero as $x \rightarrow\infty$.
  So the graph oscillates forever with ever decreasing oscillations, so has a horizontal asymptote given by $y = \int_0^\infty \sin(\pi t^2) dt$.
  A computer gives this value as $\frac{\sqrt{2}}{4}$.
  It gets to this value rather slowly, however: again the computer gives:
  $$S(10^6) = 0.3535532314.$$
  Compared with $\frac{\sqrt{2}}{4}= 0.353553391$, we see that even at $t = 10^6$, there is agreement only up to six decimal places!
• Show that the mean value of the velocity of a particle moving along the $x$-axis over a time interval $[a,b]$ is equal to the average velocity over that time interval: i.e. $A(v) = \frac{x(b) - x(a)}{b - a}$, where $v = \frac{dx}{dt}$.
  
  By definition, the mean of the velocity is:
  $$A(v) = \frac{1}{b - a} \int_a^b v(t) dt = \frac{1}{b - a} \int_a^b \frac{dx}{dt} dt$$
  $$= \frac{1}{b - a} \int_a^b d(x(t)) = \frac{1}{b - a}(x(b) -x(a)) = \frac{x(b) - x(a)}{b - a}.$$
  Here we have used $FTC$.
• Use the double angle formulas:
  $$2\cos^2(\frac{x}{2}) = 1 + \cos(x),\hspace{10pt} 2\sin^2(\frac{x}{2}) = 1 - \cos(x),$$
  to determine the integrals:
  $$I_1 = \int_0^{2\pi} \cos^2(\frac{x}{2}) dx, \hspace{10pt} I_2 = \int_0^{2\pi} \sin^2(\frac{x}{2}) dx .$$
  We first note that since $\sin^2(\frac{x}{2}) + \cos^2(\frac{x}{2}) = 1$, we have $I_1 + I_2 = \int_0^{2\pi} 1 dx = 2\pi$.
  So it suffices to do the first integral.
  We have:
  $$2I_1 = \int_0^{2\pi} 2\cos^(\frac{x}{2}) dx = \int_0^{2\pi} (1 + \cos(2x) ) dx.$$
  Now $G(x) = x + \frac{\sin(2x)}{2}$ has the derivative $1 + \cos(2x)$.
  So by $FTC$, we have:
  $$2I_1 = G(2\pi) - G(0) = 2\pi + \frac{1}{2}(\sin(4\pi) - \sin(0)) = 2\pi.$$
  So $I_1 = \pi$, so $I_2 = \pi$ also.
• Using $FTC$, calculate the mean value $A(f)$ and find $\xi$, such that $A(f) = f(\xi)$ for the following functions:
  • $f(t) = (7 - 2t)$ on $[-2,3]$.
  • $f(t) = (t^2 + t)$ on $[-1,2]$.
  • $f(t) = \frac{1}{t}$ on $[1,2]$.
  Using $FTC$, calculate the mean value $A(f)$ and find $\xi$, such that $A(f) = f(\xi)$ for the following functions and compare with your previous numerical estimates:
  • $f(t) = (7 - 2t)$ on $[-2,3]$.
    $$A(f) = \frac{1}{3 - (-2)} \int_{-2}^3 (7 - 2t) dt = \frac{1}{5} \int_{-2}^3 (7 - 2t) dt.$$
    Now $G(t) = 7t - t^2$ has the derivative $7 - 2t$, so by $FTC$, we get:
    $$A(f) = \frac{1}{5}(G(3) - G(-2)) = \frac{1}{5}(21 - 9 - (-14 - 4)) = \frac{1}{5}(30) = 6.$$
    This agrees with our previous estimate exactly.
  • $f(t) = (t^2 + t)$ on $[-1,2]$.
    $$A(f) = \frac{1}{2 - (-1)} \int_{-1}^2 (t^2 + t) dt = \frac{1}{3} \int_{-1}^2 (t^2 + t) dt.$$
    Now $G(t) = \frac{1}{3}t^3 + \frac{1}{2}t^2 = \frac{1}{6}(2t^3 + 3t^2)$ has the derivative $t^2 + t$, so by $FTC$, we get:
    $$A(f) = \frac{1}{3}(G(2) - G(-1)) = \frac{1}{18}(2(8) + 3(4) - (2(-1) + 3)) = \frac{1}{18}(27) = \frac{3}{2}.$$
    Again, this agrees with our previous estimate exactly.
  • $f(t) = \frac{1}{t}$ on $[1,2]$.
    $$A(f) = \frac{1}{2 - (1)} \int_{1}^2 \frac{1}{t} dt = \int_{1}^2 \frac{1}{t} dt.$$
    Now $G(t) = \ln(t)$ has the derivative $\frac{1}{t}$, so by $FTC$, we get:
    $$A(f) = G(2) - G(1) = \ln(2) - \ln(1) = \ln(2) = 0.693147.$$
    So our previous estimate of $0.7$ was fairly close.