Skip to main content
Mathematics LibreTexts

5.2E: Exercises for Section 5.2

  • Page ID
    53384
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    In exercises 1 - 4, express the limits as integrals.

    1) \(\displaystyle \lim_{n→∞}\sum_{i=1}^n(x^∗_i)Δx\) over \([1,3]\)

    2) \(\displaystyle \lim_{n→∞}\sum_{i=1}^n(5(x^∗_i)^2−3(x^∗_i)^3)Δx\) over \([0,2]\)

    Answer
    \(\displaystyle ∫^2_0(5x^2−3x^3)\,dx\)

    3) \(\displaystyle \lim_{n→∞}\sum_{i=1}^n\sin^2(2πx^∗_i)Δx\) over \([0,1]\)

    4) \(\displaystyle \lim_{n→∞}\sum_{i=1}^n\cos^2(2πx^∗_i)Δx\) over \([0,1]\)

    Answer
    \(\displaystyle ∫^1_0\cos^2(2πx)\,dx\)

    In exercises 5 - 10, given \(L_n\) or \(R_n\) as indicated, express their limits as \(n→∞\) as definite integrals, identifying the correct intervals.

    5) \(\displaystyle L_n=\frac{1}{n}\sum_{i=1}^n\frac{i−1}{n}\)

    6) \(\displaystyle R_n=\frac{1}{n}\sum_{i=1}^n\frac{i}{n}\)

    Answer
    \(\displaystyle ∫^1_0x\,dx\)

    7) \(\displaystyle Ln=\frac{2}{n}\sum_{i=1}^n(1+2\frac{i−1}{n})\)

    8) \(\displaystyle R_n=\frac{3}{n}\sum_{i=1}^n(3+3\frac{i}{n})\)

    Answer
    \(\displaystyle ∫^6_3x\,dx\)

    9) \(\displaystyle L_n=\frac{2π}{n}\sum_{i=1}^n2π\frac{i−1}{n}\cos(2π\frac{i−1}{n})\)

    10 \(\displaystyle R_n=\frac{1}{n}\sum_{i=1}^n(1+\frac{i}{n})\log((1+\frac{i}{n})^2)\)

    Answer
    \(\displaystyle ∫^2_1x\log(x^2)\,dx\)

    In exercises 11 - 16, evaluate the integrals of the functions graphed using the formulas for areas of triangles and circles, and subtracting the areas below the \(x\)-axis.

    11)

    A graph containing the upper half of three circles on the x axis. The first has center at (1,0) and radius one. It corresponds to the function sqrt(2x – x^2) over [0,2]. The second has center at (4,0) and radius two. It corresponds to the function sqrt(-12 + 8x – x^2) over [2,6]. The last has center at (9,0) and radius three. It corresponds to the function sqrt(-72 + 18x – x^2) over [6,12]. All three semi circles are shaded – the area under the curve and above the x axis.

    12)

    A graph of three isosceles triangles corresponding to the functions 1 - |x-1| over [0,2], 2 - |x-4| over [2,4], and 3 - |x-9| over [6,12]. The first triangle has endpoints at (0,0), (2,0), and (1,1). The second triangle has endpoints at (2,0), (6,0), and (4,2). The last has endpoints at (6,0), (12,0), and (9,3). All three are shaded.

    Answer
    \( 1+2⋅2+3⋅3=14\)

    13)

    A graph with three parts. The first is the upper half of a circle with center at (1, 0) and radius 1, which corresponds to the function sqrt(2x – x^2) over [0,2]. The second is a triangle with endpoints at (2, 0), (6, 0), and (4, -2), which corresponds to the function |x-4| - 2 over [2, 6]. The last is the upper half of a circle with center at (9, 0) and radius 3, which corresponds to the function sqrt(-72 + 18x – x^2) over [6,12]. All three are shaded.

    14)

    A graph of three shaded triangles. The first has endpoints at (0, 0), (2, 0), and (1, 1) and corresponds to the function 1 - |x-1| over [0, 2]. The second has endpoints at (2, 0), (6, 0), and (4, -2) and corresponds to the function |x-4| - 2 over [2, 6]. The third has endpoints at (6, 0), (12, 0), and (9, 3) and corresponds to the function 3 - |x-9| over [6, 12].

    Answer
    \(1−4+9=6\)

    15)

    A graph with three shaded parts. The first is the upper half of a circle with center at (1, 0) and radius one. It corresponds to the function sqrt(2x – x^2) over [0, 2]. The second is the lower half of a circle with center at (4, 0) and radius two, which corresponds to the function -sqrt(-12 + 8x – x^2) over [2, 6]. The last is the upper half of a circle with center at (9, 0) and radius three. It corresponds to the function sqrt(-72 + 18x – x^2) over [6, 12].

    16)

    A graph with three shaded parts. The first is a triangle with endpoints at (0, 0), (2, 0), and (1, 1), which corresponds to the function 1 - |x-1| over [0, 2] in quadrant 1. The second is the lower half of a circle with center at (4, 0) and radius two, which corresponds to the function –sqrt(-12 + 8x – x^2) over [2, 6]. The last is a triangle with endpoints at (6, 0), (12, 0), and (9, 3), which corresponds to the function 3 - |x-9| over [6, 12].

    Answer
    \(1−2π+9=10−2π\)

    In exercises 17 - 24, evaluate the integral using area formulas.

    17) \(\displaystyle ∫^3_0(3−x)\,dx\)

    18) \(\displaystyle ∫^3_2(3−x)\,dx\)

    Answer
    The integral is the area of the triangle, \(\frac{1}{2}.\)

    19) \(\displaystyle ∫^3_{−3}(3−|x|)\,dx\)

    20) \(\displaystyle ∫^6_0(3−|x−3|)\,dx\)

    Answer
    The integral is the area of the triangle, \(9.\)

    21) \(\displaystyle ∫^2_{−2}\sqrt{4−x^2}\,dx\)

    22) \(\displaystyle ∫^5_1\sqrt{4−(x−3)^2}\,dx\)

    Answer
    The integral is the area \(\frac{1}{2}πr^2=2π.\)

    23) \(\displaystyle ∫^{12}_0\sqrt{36−(x−6)^2}\,dx\)

    24) \(\displaystyle ∫^3_{−2}(3−|x|)\,dx\)

    Answer
    The integral is the area of the “big” triangle less the “missing” triangle, \(9−\frac{1}{2}.\)

    In exercises 25 - 28, use averages of values at the left (L) and right (R) endpoints to compute the integrals of the piecewise linear functions with graphs that pass through the given list of points over the indicated intervals.

    25) \( {(0,0),(2,1),(4,3),(5,0),(6,0),(8,3)}\) over \( [0,8]\)

    26) \({(0,2),(1,0),(3,5),(5,5),(6,2),(8,0)}\) over \([0,8]\)

    Answer
    \( L=2+0+10+5+4=21,\; R=0+10+10+2+0=22,\; \dfrac{L+R}{2}=21.5\)

    27) \( {(−4,−4),(−2,0),(0,−2),(3,3),(4,3)}\) over \( [−4,4]\)

    28) \( {(−4,0),(−2,2),(0,0),(1,2),(3,2),(4,0)}\) over \( [−4,4]\)

    Answer
    \( L=0+4+0+4+2=10,\;R=4+0+2+4+0=10,\;\dfrac{L+R}{2}=10\)

    Suppose that \(\displaystyle ∫^4_0f(x)\,dx=5\) and \(\displaystyle ∫^2_0f(x)\,dx=−3\), and \(\displaystyle ∫^4_0g(x)\,dx=−1\) and \(\displaystyle ∫^2_0g(x)\,dx=2\). In exercises 29 - 34, compute the integrals.

    29) \(\displaystyle ∫^4_0(f(x)+g(x))\,dx\)

    30) \(\displaystyle ∫^4_2(f(x)+g(x))\,dx\)

    Answer
    \(\displaystyle ∫^4_2f(x)\,dx+∫^4_2g(x)\,dx=8−3=5\)

    31) \(\displaystyle ∫^2_0(f(x)−g(x))\,dx\)

    32) \(\displaystyle ∫^4_2(f(x)−g(x))\,dx\)

    Answer
    \(\displaystyle ∫^4_2f(x)\,dx−∫^4_2g(x)\,dx=8+3=11\)

    33) \(\displaystyle ∫^2_0(3f(x)−4g(x))\,dx\)

    34) \(\displaystyle ∫^4_2(4f(x)−3g(x))\,dx\)

    Answer
    \(\displaystyle 4∫^4_2f(x)\,dx−3∫^4_2g(x)\,dx=32+9=41\)

    In exercises 35 - 38, use the identity \(\displaystyle ∫^A_{−A}f(x)\,dx=∫^0_{−A}f(x)\,dx+∫^A_0f(x)\,dx\) to compute the integrals.

    35) \(\displaystyle ∫^π_{−π}\frac{\sin t}{1+t^2}dt\) (Hint: \(\displaystyle \sin(−t)=−\sin(t))\)

    36) \(\displaystyle ∫^{\sqrt{π}}_\sqrt{−π}\frac{t}{1+\cos t}dt\)

    Answer
    The integrand is odd; the integral is zero.

    37) \(\displaystyle ∫^3_1(2−x)\,dx\) (Hint: Look at the graph of \(f\).)

    38) \(\displaystyle ∫^4_2(x−3)^3\,dx\) (Hint: Look at the graph of \(f\).)

    Answer
    The integrand is antisymmetric with respect to \(x=3.\) The integral is zero.

    In exercises 39 - 44, given that \(\displaystyle ∫^1_0x\,dx=\frac{1}{2},\;∫^1_0x^2\,dx=\frac{1}{3},\) and \(\displaystyle ∫^1_0x^3\,dx=\frac{1}{4}\), compute the integrals.

    39) \(\displaystyle ∫^1_0(1+x+x^2+x^3)\,dx\)

    40) \(\displaystyle ∫^1_0(1−x+x^2−x^3)\,dx\)

    Answer
    \(\displaystyle 1−\frac{1}{2}+\frac{1}{3}−\frac{1}{4}=\frac{7}{12}\)

    41) \(\displaystyle ∫^1_0(1−x)^2\,dx\)

    42) \(\displaystyle ∫^1_0(1−2x)^3\,dx\)

    Answer
    \(\displaystyle ∫^1_0(1−6x+12x^2−8x^3)\,dx=1−6\left( \frac{1}{2} \right)+12\left(\frac{1}{3}\right)−8\left(\frac{1}{4}\right)=1-3+4-2=0\)

    43) \(\displaystyle ∫^1_0\left(6x−\tfrac{4}{3}x^2\right)\,dx\)

    44) \(\displaystyle ∫^1_0(7−5x^3)\,dx\)

    Answer
    \(7−\frac{5}{4}=\frac{23}{4}\)

    In exercises 45 - 50, use the comparison theorem.

    45) Show that \(\displaystyle ∫^3_0(x^2−6x+9)\,dx≥0.\)

    46) Show that \(\displaystyle ∫^3_{−2}(x−3)(x+2)\,dx≤0.\)

    Answer
    The integrand is negative over \([−2,3].\)

    47) Show that \(\displaystyle ∫^1_0\sqrt{1+x^3}\,dx≤∫^1_0\sqrt{1+x^2}\,dx\).

    48) Show that \(\displaystyle ∫^2_1\sqrt{1+x}\,dx≤∫^2_1\sqrt{1+x^2}\,dx.\)

    Answer
    \(x≤x^2\) over \([1,2]\), so \(\sqrt{1+x}≤\sqrt{1+x^2}\) over \([1,2].\)

    49) Show that \(\displaystyle ∫^{π/2}_0\sin tdt≥\frac{π}{4}\) (Hint: \(\sin t≥\frac{2t}{π}\) over \( [0,\frac{π}{2}])\)

    50) Show that \(\displaystyle ∫^{π/4}_{−π/4}\cos t\,dt≥π\sqrt{2}/4\).

    Answer
    \(\cos(t)≥\dfrac{\sqrt{2}}{2}\). Multiply by the length of the interval to get the inequality.

    In exercises 51 - 56, find the average value \(f_{ave}\) of \(f\) between \(a\) and \(b\), and find a point \(c\), where \(f(c)=f_{ave}\)

    51) \( f(x)=x^2,\; a=−1,\; b=1\)

    52) \( f(x)=x^5,\; a=−1,\; b=1\)

    Answer
    \(f_{ave}=0;\; c=0\)

    53) \( f(x)=\sqrt{4−x^2},\; a=0,\; b=2\)

    54) \(f(x)=3−|x|,\; a=−3,\; b=3\)

    Answer
    \(\frac{3}{2}\) when \(c=±\frac{3}{2}\)

    55) \(f(x)=\sin x,\; a=0,\; b=2π\)

    56) \( f(x)=\cos x,\; a=0,\; b=2π\)

    Answer
    \(f_{ave}=0;\; c=\dfrac{π}{2},\; \dfrac{3π}{2}\)

    In exercises 57 - 60, approximate the average value using Riemann sums \(L_{100}\) and \(R_{100}\). How does your answer compare with the exact given answer?

    57) [T] \(y=\ln(x)\) over the interval \( [1,4]\); the exact solution is \(\dfrac{\ln(256)}{3}−1.\)

    58) [T] \(y=e^{x/2}\) over the interval \([0,1]\); the exact solution is \( 2(\sqrt{e}−1).\)

    Answer
    \(L_{100}=1.294,\; R_{100}=1.301;\) the exact average is between these values.

    59) [T] \(y=\tan x\) over the interval \([0,\frac{π}{4}]\); the exact solution is \(\dfrac{2\ln(2)}{π}\).

    60) [T] \(y=\dfrac{x+1}{\sqrt{4−x^2}}\) over the interval \([−1,1]\); the exact solution is \(\dfrac{π}{6}\).

    Answer
    \(L_{100}×(\dfrac{1}{2})=0.5178,\; R_{100}×(\dfrac{1}{2})=0.5294\)

    In exercises 61 - 64, compute the average value using the left Riemann sums \(L_N\) for \(N=1,10,100\). How does the accuracy compare with the given exact value?

    61) [T] \(y=x^2−4\) over the interval \([0,2]\); the exact solution is \(−\frac{8}{3}\).

    62) [T] \(y=xe^{x^2}\) over the interval \([0,2]\); the exact solution is \(\frac{1}{4}(e^4−1).\)

    Answer
    \(L_1=0,\; L_{10}×(\frac{1}{2})=8.743493,\; L_{100}×(\frac{1}{2})=12.861728.\) The exact answer \(≈26.799,\) so \(L_{100}\) is not accurate.

    63) [T] \(y=\left(\dfrac{1}{2}\right)^x\) over the interval \([0,4]\); the exact solution is \(\dfrac{15}{64\ln(2)}\).

    64) [T] \( y=x\sin(x^2)\) over the interval \( [−π,0]\); the exact solution is \( \dfrac{\cos(π^2)−1}{2π.}\)

    Answer
    \(L_1×(\frac{1}{π})=1.352,L_{10}×(\frac{1}{π})=−0.1837,L_{100}×(1π)=−0.2956.\) The exact answer \(≈−0.303,\) so \(L_{100}\) is not accurate to the first decimal.

    65) Suppose that \(\displaystyle A=∫^{2π}_0\sin^2t\,dt\) and \(\displaystyle B=∫^{2π}_0\cos^2t\,dt.\) Show that \(A+B=2π\) and \(A=B.\)

    66) Suppose that \(\displaystyle A=∫^{π/4}_{−π/4}\sec^2 t\,dt=π\) and \(\displaystyle B=∫^{π/4}_{−π/}4\tan^2 t\,dt.\) Show that \(A−B=\dfrac{π}{2}\).

    Answer
    Use \(\tan^2 θ+1=\sec^2 θ.\) Then, \(\displaystyle B−A=∫^{π/4}_{−π/4}1\,dx=\frac{π}{2}.\)

    67) Show that the average value of \(\sin^2 t\) over \([0,2π]\) is equal to \(1/2.\) Without further calculation, determine whether the average value of \(\sin^2 t\) over \([0,π]\) is also equal to \(1/2.\)

    68) Show that the average value of \(\cos^2 t\) over \([0,2π]\) is equal to \(1/2.\) Without further calculation, determine whether the average value of \(\cos^2(t)\) over \([0,π]\) is also equal to \(1/2.\)

    Answer
    \(\displaystyle ∫^{2π}_0\cos^2t\,dt=π,\) so divide by the length \(2π\) of the interval. \(\cos^2t\) has period \(π\), so yes, it is true.

    69) Explain why the graphs of a quadratic function (parabola) \(p(x)\) and a linear function \(ℓ(x)\) can intersect in at most two points. Suppose that \(p(a)=ℓ(a)\) and \(p(b)=ℓ(b)\), and that \(\displaystyle ∫^b_ap(t)\,dt>∫^b_aℓ(t)\,dt\). Explain why \(\displaystyle ∫^d_cp(t)\,dt >∫^d_cℓ(t)\,dt\) whenever \( a≤c<d≤b.\)

    70) Suppose that parabola \(p(x)=ax^2+bx+c\) opens downward \((a<0)\) and has a vertex of \(y=\dfrac{−b}{2a}>0\). For which interval \([A,B]\) is \(\displaystyle ∫^B_A(ax^2+bx+c)\,dx\) as large as possible?

    Answer
    The integral is maximized when one uses the largest interval on which \(p\) is nonnegative. Thus, \(A=\frac{−b−\sqrt{b^2−4ac}}{2a}\) and \(B=\frac{−b+\sqrt{b^2−4ac}}{2a}.\)

    71) Suppose \([a,b]\) can be subdivided into subintervals \(a=a_0<a_1<a_2<⋯<a_N=b\) such that either \(f≥0\) over \([a_{i−1},a_i]\) or \(f≤0\) over \([a_{i−1},a_i]\). Set \(\displaystyle A_i=∫^{a_i}_{a_{i−1}}f(t)\,dt.\)

    a. Explain why \(\displaystyle ∫^b_af(t)\,dt=A_1+A_2+⋯+A_N.\)

    b. Then, explain why \(\displaystyle ∫^b_af(t)\,dt≤∫^b_a|f(t)|\,dt.\)

    72) Suppose \(f\) and \(g\) are continuous functions such that \(\displaystyle ∫^d_cf(t)\,dt≤∫^d_cg(t)\,dt\) for every subinterval \([c,d]\) of \([a,b]\). Explain why \( f(x)≤g(x)\) for all values of \(x.\)

    Answer
    If \(f(t_0)>g(t_0)\) for some \(t_0∈[a,b]\), then since \(f−g\) is continuous, there is an interval containing \(t_0\) such that \( f(t)>g(t)\) over the interval \([c,d]\), and then \(\displaystyle ∫^d_df(t)\,dt>∫^d_cg(t)\,dt\)over this interval.

    73) Suppose the average value of \(f\) over \([a,b]\) is \(1\) and the average value of \(f\) over \([b,c]\) is \(1\) where \(a<c<b\). Show that the average value of \(f\) over \([a,c]\) is also \(1.\)

    74) Suppose that \([a,b]\) can be partitioned. taking \(a=a_0<a_1<⋯<a_N=b\) such that the average value of \(f\) over each subinterval \([a_{i−1},a_i]=1\) is equal to 1 for each \( i=1,…,N\). Explain why the average value of f over \( [a,b]\) is also equal to \(1.\)

    Answer
    The integral of f over an interval is the same as the integral of the average of f over that interval. Thus, \(\displaystyle ∫^b_af(t)\,dt=∫^{a_1}_{a_0}f(t)\,dt+∫^{a_2}_a{1_f}(t)\,dt+⋯+∫^{a_N}_{a_{N+1}}f(t)\,dt=∫^{a_1}_{a_0}1\,dt+∫^{a_2}_{a_1}1\,dt+⋯+∫^{a_N}_{a_{N+1}}1\,dt\)
    \( =(a_1−a_0)+(a_2−a_1)+⋯+(a_N−a_{N−1})=a_N−a_0=b−a\).
    Dividing through by \(b−a\) gives the desired identity.

    75) Suppose that for each \(i\) such that \( 1≤i≤N\) one has \(\displaystyle ∫^i_{i−1}f(t)\,dt=i\). Show that \(\displaystyle ∫^N_0f(t)\,dt=\frac{N(N+1)}{2}.\)

    76) Suppose that for each \(i\) such that \(1≤i≤N\) one has \(\displaystyle ∫^i_{i−1}f(t)\,dt=i^2\). Show that \(\displaystyle ∫^N_0f(t)\,dt=\frac{N(N+1)(2N+1)}{6}\).

    Answer
    \(\displaystyle ∫^N_0f(t)\,dt=\sum_{i=1}^N∫^i_{i−1}f(t)\,dt=\sum_{i=1}^Ni^2=\frac{N(N+1)(2N+1)}{6}\)

    77) [T] Compute the left and right Riemann sums \(\displaystyle L_{10}\) and \(R_{10}\) and their average \(\dfrac{L_{10}+R_{10}}{2}\) for \( f(t)=t^2\)over \( [0,1]\). Given that \(\displaystyle ∫^1_0t^2\,dt=1/3\), to how many decimal places is \( \dfrac{L_{10}+R_{10}}{2}\) accurate?

    78) [T] Compute the left and right Riemann sums, \(L_10\) and \(R_{10}\), and their average \(\dfrac{L_{10}+R_{10}}{2}\) for \( f(t)=(4−t^2)\) over \([1,2]\). Given that \(\displaystyle ∫^2_1(4−t^2)\,dt=1.66\), to how many decimal places is \(\dfrac{L_{10}+R_{10}}{2}\) accurate?

    Answer
    \( L_{10}=1.815,\;R_{10}=1.515,\;\frac{L_{10}+R_{10}}{2}=1.665,\) so the estimate is accurate to two decimal places.

    79) If \(\displaystyle ∫^5_1\sqrt{1+t^4}\,dt=41.7133...,\) what is \(\displaystyle ∫^5_1\sqrt{1+u^4}\,du?\)

    80) Estimate \(\displaystyle ∫^1_0t\,dt\) using the left and right endpoint sums, each with a single rectangle. How does the average of these left and right endpoint sums compare with the actual value \(\displaystyle ∫^1_0t\,dt?\)

    Answer
    The average is \(1/2,\) which is equal to the integral in this case.

    81) Estimate \(\displaystyle ∫^1_0t\,dt\) by comparison with the area of a single rectangle with height equal to the value of \(t\) at the midpoint \(t=\dfrac{1}{2}\). How does this midpoint estimate compare with the actual value \(\displaystyle ∫^1_0t\,dt?\)

    82) From the graph of \(\sin(2πx)\) shown:

    a. Explain why \(\displaystyle ∫^1_0\sin(2πt)\,dt=0.\)

    b. Explain why, in general, \(\displaystyle ∫^{a+1}_a\sin(2πt)\,dt=0\) for any value of \(a\).

    Answer

    A graph of the function f(x) = sin(2pi*x) over [0, 2]. The function is shaded over [.7, 1] above the curve and below to x axis, over [1,1.5] under the curve and above the x axis, and over [1.5, 1.7] above the curve and under the x axis. The graph is antisymmetric with respect o t = ½ over [0,1].

    a. The graph is antisymmetric with respect to \(t=\frac{1}{2}\) over \([0,1]\), so the average value is zero.
    b. For any value of \(a\), the graph between \([a,a+1]\) is a shift of the graph over \([0,1]\), so the net areas above and below the axis do not change and the average remains zero.

    83) If f is 1-periodic \((f(t+1)=f(t))\), odd, and integrable over \([0,1]\), is it always true that \(\displaystyle ∫^1_0f(t)\,dt=0?\)

    84) If f is 1-periodic and \(\displaystyle ∫10f(t)\,dt=A,\) is it necessarily true that \(\displaystyle ∫^{1+a}_af(t)\,dt=A\) for all \(A\)?

    Answer
    Yes, the integral over any interval of length 1 is the same.

    5.2E: Exercises for Section 5.2 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?