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Mathematics LibreTexts

5.5E and 5.6E U-Substitution Exercises

  • Page ID
    14694
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    5.5: Substitution

    In exercises 1 - 16, find the antiderivative.

    1)    \(\displaystyle∫(x+1)^4\,dx\)

    Answer:
    \(\displaystyle∫(x+1)^4\,dx \quad\) \(=\quad \displaystyle\frac{1}{5}(x+1)^5+C\)

    2)    \(\displaystyle∫(x−1)^5\,dx\)

    3)    \(\displaystyle∫(2x−3)^{−7}\,dx\)

    Answer:
    \(\displaystyle∫(2x−3)^{−7}\,dx\quad\) \(=\quad\displaystyle−\frac{1}{12(2x-3)^6}+C\)

    4) \(\displaystyle∫(3x−2)^{−11}\,dx\)

    5) \(\displaystyle∫\frac{x}{\sqrt{x^2+1}}\,dx\)

    Answer:
    \(\displaystyle∫\frac{x}{\sqrt{x^2+1}}\,dx\quad\) \(=\quad \displaystyle\sqrt{x^2+1}+C\)

    6) \(\displaystyle∫\frac{x}{\sqrt{1−x^2}}\,dx\)

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

    Answer:
    \(\displaystyle∫(x−1)(x^2−2x)^3\,dx\quad\) \(=\quad\displaystyle\frac{1}{8}(x^2−2x)^4+C\)

    8) \(\displaystyle∫(x^2−2x)(x^3−3x^2)^2\,dx\)

    9) \(\displaystyle\int\cos^3 θ\,dθ\)      (Hint: \(\cos^2 θ=1−\sin^2 θ\))

    Answer:
    \(\displaystyle\int\cos^3 θ\,dθ\quad\) \(=\quad\displaystyle \sin θ−\frac{\sin^3 θ}{3}+C\)

    10) \(\displaystyle\int\sin^3 θ\,dθ\)        (Hint: \(\sin^2 θ=1−\cos^2 θ\))

    11) \(\displaystyle\int x(1−x)^{99}\,dx\)

    Answer:
    \(\displaystyle\int x(1−x)^{99}\,dx\quad\) \(=\quad\displaystyle\frac{(1−x)^{101}}{101}−\frac{(1−x)^{100}}{100}+C\)

    12) \(\displaystyle∫t(1−t^2)^{10}\,dt\)

    13) \(\displaystyle∫(11x−7)^{−3}\,dx\)

    Answer:
    \(\displaystyle∫(11x−7)^{−3}\,dx \quad = \quad \displaystyle−\frac{1}{22(11x−7)^2}+C\)

    14) \(\displaystyle∫(7x−11)^4\,dx\)

    15) \(\displaystyle\int\cos^3 θ\sin θ\,dθ\)

    Answer:
    \(\displaystyle\int\cos^3 θ\sin θ\,dθ \quad = \quad −\frac{\cos^4 θ}{4}+C\)

    16) \(\displaystyle∫\frac{x^2}{(x^3−3)^2}\,dx\)

    Answer:
    \(\displaystyle∫\frac{x^2}{(x^3−3)^2}\,dx \quad = \quad −\frac{1}{3(x^3−3)}+C\)

    u-Substitution with Definite Integrals

    In exercises 17 - 22, evaluate the definite integral.

    17) \(\displaystyle∫^1_0x\sqrt{1−x^2}\,dx\)

    18) \(\displaystyle∫^1_0\frac{x}{\sqrt{1+x^2}}\,dx\)

    Answer:
    \(\displaystyle u=1+x^2,\quad du=2x\,dx,\quad ∫^1_0\frac{x}{\sqrt{1+x^2}}\,dx = \frac{1}{2}∫^2_1u^{−1/2}du=\sqrt{2}−1\)

    19) \(\displaystyle∫^2_0\frac{t}{\sqrt{5+t^2}}\,dt\)

    20) \(\displaystyle∫^1_0\frac{t}{\sqrt{1+t^3}}\,dt\)

    Answer:
    \(\displaystyle u=1+t^3,\quad du=3t^2,\quad ∫^1_0\frac{t}{\sqrt{1+t^3}}\,dt = \frac{1}{3}∫^2_1u^{−1/2}du=\frac{2}{3}(\sqrt{2}−1)\)

    21) \(\displaystyle\int^{π/4}_0\sec^2 θ\tan θ\,dθ\)

    22) \(\displaystyle\int^{π/4}_0\frac{\sin θ}{\cos^4 θ}\,dθ\)

    Answer:
    \(\displaystyle u=\cos θ,\quad du=−\sin θ\,dθ,\quad \int^{π/4}_0\frac{\sin θ}{\cos^4 θ}\,dθ = -∫_1^{\sqrt{2}/2}u^{−4}\,du = ∫^1_{\sqrt{2}/2}u^{−4}\,du=\frac{1}{3}(2\sqrt{2}−1)\)

    u-Substitution with a Twist

    In exercises 23 - 28, find the antiderivative.  Then check your answer by showing its derivative can be simplified to the original integrand.

    23)  \(\displaystyle ∫x\sqrt{x+1}\,dx\)

    Answer:
    \(\displaystyle ∫x\sqrt{x+1}\,dx \quad = \quad \frac{2}{15}(x+1)^{3/2}(3x−2)+C\)

    24)  \(\displaystyle \int\frac{x}{\sqrt{3x+1}}\,dx\)

    Answer:
    \(\displaystyle \int\frac{x}{\sqrt{3x+1}}\,dx \quad = \quad \frac{2}{27}\sqrt{3x+1}\big(3x - 2\big) + C\)

    25)  \(\displaystyle \int\frac{x+2}{(x-1)^{3/2}}\,dx\)

    Answer:
    \(\displaystyle \int\frac{x+2}{(x-1)^{3/2}}\,dx \quad = \quad \frac{2(x - 4)}{\sqrt{x-1}} + C\)

    Check (using quotient rule):
    \( \begin{align*} \dfrac{d}{dx}\left( \frac{2(x - 4)}{\sqrt{x-1}} + C \right) \quad &= \quad \dfrac{2\sqrt{x-1} - 2(x - 4)\cdot\frac{1}{2}(x-1)^{-1/2}}{x-1} \\[5pt]
    &=\dfrac{(x-1)^{-1/2}\big( 2(x-1) - (x - 4)  \big)}{x-1}  \\[5pt]
    &=\dfrac{2x - 2 - x + 4}{(x-1)^{3/2}} \\[5pt]
    &=\dfrac{x+2}{(x-1)^{3/2}}  \quad \checkmark \end{align*}\)

    26)  \(\displaystyle \int t^2\sqrt{3 - t} \,dt\)

    Answer:
    \(\displaystyle \begin{align*} \int t^2\sqrt{3 - t} \,dt \quad &=  -6(3 - t)^{3/2} + \frac{12}{5}(3 - t)^{5/2} - \frac{2}{7}(3 - t)^{7/2} + C \\[5pt]
    &= -\frac{2}{35}(3 - t)^{3/2}\big[5t^2 + 12t + 24\big] + C \end{align*}\)

    Check (using product rule):
    \( \begin{align*} \dfrac{d}{dx}\left( -\frac{2}{35}(3 - t)^{3/2}\big[5t^2 + 12t + 24\big] + C \right) &=-\frac{2}{35}\bigg[ (3 - t)^{3/2}(10t + 12) - \frac{3}{2}(3 - t)^{1/2}\big[5t^2 + 12t + 24\big]\bigg] \\[5pt]
    &=-\frac{1}{35}(3 - t)^{1/2}\bigg[ 2(3 - t)(10t + 12) - 3\big[5t^2 + 12t + 24\big]\bigg] \\[5pt]
    &=-\frac{1}{35}(3 - t)^{1/2}\big[ 4(3 - t)(5t + 6) - 15t^2 - 36t - 72\big] \\[5pt]
    &=-\frac{1}{35}(3 - t)^{1/2}\big[ 4(18 + 9t - 5t^2) - 15t^2 - 36t - 72\big] \\[5pt]
    &=-\frac{1}{35}(3 - t)^{1/2}\big[ 72 + 36t - 20t^2 - 15t^2 - 36t - 72\big] \\[5pt]
    &=-\frac{1}{35}(3 - t)^{1/2}\big[ -35t^2\big] \\[5pt]
    &=t^2\sqrt{3 - t}  \quad \checkmark \end{align*}\)

    27)  \(\displaystyle \int t^3\,\sqrt{t^2 + 5} \,dt\)

    Answer:
    \(\displaystyle \begin{align*} \int t^3\,\sqrt{t^2 + 5} \,dt \quad &=  \frac{1}{2}\int (u - 5)u^{1/2}\, du \\[5pt]
    &= \frac{(t^2 + 5)^{5/2}}{5} - \frac{5}{3}(t^2 + 5)^{3/2} + C\\[5pt]
    &= \frac{(t^2 + 5)^{3/2}}{15}\big[ 3t^2 - 10 \big] + C \end{align*}\)

    Check (using product rule):
    \( \begin{align*} \dfrac{d}{dx}\left( \frac{(t^2 + 5)^{3/2}}{15}\big[ 3t^2 - 10 \big] + C \right) &=\frac{6t(t^2 + 5)^{3/2}}{15} + (3t^2 - 10)\cdot\frac{3}{2}\frac{(t^2 + 5)^{1/2}}{15}\cdot 2t \\[5pt]
    &=\frac{3t(t^2+5)^{1/2}}{15}\big[ 2(t^2+5) + (3t^2-10) \big] \\[5pt]
    &=\frac{t}{5}\sqrt{t^2 + 5} \big[ 5t^2 \big] \\[5pt]
    &=t^3\,\sqrt{t^2 + 5} \quad \checkmark \end{align*}\)

    28)   \(\displaystyle \int x\sqrt[3]{x - 2} \,dx\)

    Answer:
    \(\displaystyle \begin{align*} \int x\sqrt[3]{x - 2} \,dx \quad &=  \frac{3}{7}(x-2)^{7/3}+\frac{3}{2}(x-2)^{4/3}+ C \\[5pt]
    &= \frac{3}{14}(x-2)^{4/3}[2x+3] + C \end{align*}\)

    Check (using product rule):
    \( \begin{align*} \dfrac{d}{dx}\left( \frac{3}{14}(x-2)^{4/3}[2x+3] + C \right) &=\frac{3}{7}(x-2)^{4/3} + (2x+3)\cdot\frac{3}{14}\cdot\frac{4}{3}(x-2)^{1/3} \\[5pt]
    &=\frac{3}{7}(x-2)^{4/3} + \frac{2}{7}(x-2)^{1/3}(2x+3) \\[5pt]
    &=\frac{1}{7}(x-2)^{1/3}\big[ 3(x-2) + 2(2x+3) \big] \\[5pt]
    &=\frac{1}{7}(x-2)^{1/3}\big[ 3x-6 + 4x+6 \big] \\[5pt]
    &=\frac{1}{7}(x-2)^{1/3}\big[ 7x \big] \\[5pt]
    &=x\sqrt[3]{x - 2} \quad \checkmark \end{align*}\)

     

     

    5.6: Integrals Involving Exponential and Logarithmic Functions

    For exercises 1 - 8, compute each indefinite integral.

    1)     \(\displaystyle ∫e^{2x}\,dx\)

    2)    \(\displaystyle ∫e^{−3x}\,dx\)

    Answer:
    \(\displaystyle ∫e^{−3x}\,dx \quad = \quad \frac{−1}{3}e^{−3x}+C\)

    3)     \(\displaystyle ∫2^x\,dx\)

    4)     \(\displaystyle ∫3^{−x}\,dx\)

    Answer:
    \(\displaystyle ∫3^{−x}\,dx \quad = \quad −\frac{3^{−x}}{\ln 3}+C\)

    5)     \(\displaystyle ∫\frac{1}{2x}\,dx\)

    6)     \(\displaystyle ∫\frac{2}{x}\,dx\)

    Answer:
    \(\displaystyle ∫\frac{2}{x}\,dx \quad = \quad 2\ln x+C \quad = \quad \ln(x^2)+C\)

    7)     \(\displaystyle ∫\frac{1}{x^2}\,dx\)

    8)     \(\displaystyle ∫\frac{1}{\sqrt{x}}\,dx\)

    Answer:
    \(\displaystyle ∫\frac{1}{\sqrt{x}}\,dx \quad = \quad  2\sqrt{x}+C\)

    In exercises 9 - 16, find each indefinite integral by using appropriate substitutions.

    9)  \(\displaystyle ∫\frac{\ln x}{x}\,dx\)

    10)  \(\displaystyle ∫\frac{dx}{x(\ln x)^2}\)

    Answer:
    \(\displaystyle ∫\frac{dx}{x(\ln x)^2} \quad = \quad −\frac{1}{\ln x}+C\)

    11) \(\displaystyle ∫xe^{−x^2}\,dx\)

    12) \(\displaystyle ∫x^2e^{−x^3}\,dx\)

    Answer:
    \(\displaystyle ∫x^2e^{−x^3}\,dx \quad = \quad \dfrac{−e^{−x^3}}{3}+C\)

    13) \(\displaystyle ∫e^{\sin x}\cos x\,dx\)

    14) \(\displaystyle ∫e^{\tan x}\sec^2 x\,dx\)

    Answer:
    \(\displaystyle ∫e^{\tan x}\sec^2 x\,dx\quad = \quad e^{\tan x}+C\)

    15) \(\displaystyle ∫\frac{e^{\ln x}}{x}\,dx\)

    16) \(\displaystyle ∫\frac{e^{\ln(1−t)}}{1−t}\,dt\)

    Answer:
    \(\displaystyle ∫\frac{e^{\ln(1−t)}}{1−t}\,dt = \int \frac{1-t}{1-t}\,dt = \int 1\, dt \quad = \quad t+C\)
     

    More u-Substitutions with Definite Integrals

    In exercises 17 - 21, evaluate the definite integral.

    17)  \(\displaystyle ∫^2_1\frac{1+2x+x^2}{3x+3x^2+x^3}\,dx\)

    Answer:
    \(\displaystyle ∫^2_1\frac{1+2x+x^2}{3x+3x^2+x^3}\,dx \quad = \quad \frac{1}{3}\ln(\frac{26}{7})\)

    18)  \(\displaystyle ∫^{π/4}_0\tan x\,dx\)

    19)  \(\displaystyle ∫^{π/3}_0\frac{\sin x−\cos x}{\sin x+\cos x}\,dx\)

    Answer:
    \(\displaystyle ∫^{π/3}_0\frac{\sin x−\cos x}{\sin x+\cos x}\,dx \quad = \quad \ln(\sqrt{3}−1)\)

    20)  \(\displaystyle ∫^{π/2}_{π/6}\csc x\,dx\)

    21)  \(\displaystyle ∫^{π/3}_{π/4}\cot x\,dx\)

    Answer:
    \(\displaystyle ∫^{π/3}_{π/4}\cot x\,dx \quad = \quad \frac{1}{2}\ln\frac{3}{2}\)

    Some Interesting u-Substitutions

    In exercises 22 - 29, integrate using the indicated substitution.

    22) \(\displaystyle ∫\frac{x}{x−100}\,dx;\quad u=x−100\)

    23) \(\displaystyle ∫\frac{y−1}{y+1}\,dy;\quad u=y+1\)

    Answer:
    \( \displaystyle ∫\frac{y−1}{y+1}\,dy \quad = \quad y−2\ln|y+1|+C\)

    24) \(\displaystyle ∫\frac{1−x^2}{3x−x^3}\,dx;\quad u=3x−x^3\)

    25) \(\displaystyle ∫\frac{\sin x+\cos x}{\sin x−\cos x}\,dx;\quad u=\sin x−\cos x\)

    Answer:
    \(\displaystyle ∫\frac{\sin x+\cos x}{\sin x−\cos x}\,dx \quad=\quad \ln|\sin x−\cos x|+C\)

    26) \(\displaystyle ∫e^{2x}\sqrt{1−e^{2x}}\,dx;\quad u=e^{2x}\)

    27) \(\displaystyle ∫\ln(x)\frac{\sqrt{1−(\ln x)^2}}{x}\,dx;\quad u=\ln x\)

    Answer:
    \(\displaystyle ∫\ln(x)\frac{\sqrt{1−(\ln x)^2}}{x}\,dx \quad = \quad −\frac{1}{3}(1−(\ln x^2))^{3/2}+C\)

    28)   \(\displaystyle \int \frac{\sqrt{x}}{\sqrt{x} + 2}\,dx; \quad u = \sqrt{x} + 2\)

    Answer:
    \(\displaystyle \int \frac{\sqrt{x}}{\sqrt{x} + 2}\,dx \quad = \quad \left( \sqrt{x} + 2 \right)^2 - 8\left( \sqrt{x} + 2 \right) + 8\ln\left( \sqrt{x} + 2 \right) + C\)

    29)   \(\displaystyle \int e^x\sec(e^x+1)\tan(e^x+1)\,dx; \quad u = e^{x} + 1\)

    Answer:
    \(\displaystyle \int e^x\sec(e^x+1)\tan(e^x+1)\,dx \quad = \quad \sec(e^x+1) + C\)

     

    In exercises 29 - 35, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms.

    30) \(\displaystyle ∫\tan(2x)\,dx\)

    Answer:
    \( \displaystyle ∫\tan(2x)\,dx \quad = \quad −\frac{1}{2}\ln|\cos(2x)| + C \quad = \quad \frac{1}{2}\ln|\sec(2x)| + C\)
    Solution:
    \( \begin{align*} \displaystyle ∫\tan(2x)\,dx  &= ∫\frac{\sin(2x)}{\cos(2x)}\,dx \\[5pt]
    &= -\frac{1}{2}\int\frac{1}{u}\,du & & \text{Letting } u = \cos(2x), \, \text{and}\, du = -2\sin(2x) dx \\[5pt]
    &= -\frac{1}{2}\ln|u| + C & & \text{Integrating in terms of}\, u \\[5pt]
    &=−\frac{1}{2}\ln|\cos(2x)| + C \quad = \quad \frac{1}{2}\ln|\sec(2x)| + C & & \text{Going back to}\,x \\[5pt]
    & & & \text{Gives us two equivalent forms of the antiderivative} \end{align*} \)

    31) \(\displaystyle ∫\sec 5x\,dx\)

    Answer:
    \( \displaystyle ∫\sec 5x\,dx \quad = \quad \frac{1}{5}\ln|\tan 5x + \sec 5x | + C \)
    Solution:
    \( \begin{align*} \displaystyle ∫\sec 5x\,dx  &= ∫\sec 5x\cdot\frac{\sec 5x + \tan 5x}{\sec 5x + \tan 5x}\,dx \\[5pt]
    &= \frac{1}{5} ∫\frac{5(\sec^2 5x + \sec 5x\tan 5x)}{\tan 5x + \sec 5x}\,dx \\[5pt]
    &= \frac{1}{5}\ln|\tan 5x + \sec 5x | + C  \end{align*} \)

    32) \(\displaystyle ∫\frac{x\sin(x^2)}{\cos(x^2)}\,dx\)

    33) \(\displaystyle ∫\frac{\sin(3x)−\cos(3x)}{\sin(3x)+\cos(3x)}\,dx\)

    Answer:
    \( \displaystyle ∫\frac{\sin(3x)−\cos(3x)}{\sin(3x)+\cos(3x)}\,dx \quad = \quad −\frac{1}{3}\ln|\sin(3x)+\cos(3x)| + C\)

    34) \(\displaystyle ∫x\csc(x^2)\,dx\)

    Answer:
    \( \displaystyle ∫x\csc(x^2)\,dx \quad = \quad −\frac{1}{2}\ln∣\csc(x^2)+\cot(x^2)∣+C\)

    35) \(\displaystyle ∫\frac{e^x−e^{−x}}{e^x+e^{−x}}\,dx\)

    Answer:
    \( \displaystyle ∫\frac{e^x−e^{−x}}{e^x+e^{−x}}\,dx \quad = \quad \ln∣e^x+e^{−x}∣+C \quad = \quad \ln(e^x+e^{−x})+C\)

    36) \(\displaystyle ∫\ln(\cos x)\tan x\,dx\)

    37) \(\displaystyle ∫\ln(\csc x)\cot x\,dx\)

    Answer:
    \( \displaystyle ∫\ln(\csc x)\cot x\,dx \quad = \quad −\frac{1}{2}(\ln(\csc x))^2+C\)

     

    In exercises 38 - 39, \(f(x)≥0\) for \(a≤x≤b\). Find the area under the graph of \(f(x)\) between the given values \(a\) and \(b\) by integrating.

    38) \(f(x)=2^{−x};\quad a=1,b=2\)

    39) \(f(x)=2^{−x};\quad a=3,b=4\)

    Answer:
    \(\dfrac{1}{\ln(65,536)}\)

     

    40) Find the area under the graph of the function \( f(x)=xe^{−x^2}\) between \(x=0\) and \(x=5\).

    41) Compute the integral of \(f(x)=xe^{−x^2}\) and find the smallest value of \(N\) such that the area under the graph \(f(x)=xe^{−x^2}\) between \( x=N\) and \(x=N+10\) is, at most, \(0.01\).

    Answer:
    \(\displaystyle ∫^{N+1}_Nxe^{−x^2}\,dx=\frac{1}{2}(e^{−N^2}−e^{−(N+1)^2}).\) The quantity is less than \(0.01\) when \(N=2\).

    42) Find the limit, as \(N\) tends to infinity, of the area under the graph of \(f(x)=xe^{−x^2}\) between \(x=0\) and \(x=5\).

    43) Show that \(\displaystyle ∫^b_a\frac{dt}{t}=∫^{1/a}_{1/b}\frac{dt}{t}\) when \(0<a≤b\).

    Answer:
    \(\displaystyle ∫^b_a\frac{dx}{x}=\ln(b)−\ln(a)=\ln(\frac{1}{a})−\ln(\frac{1}{b})=∫^{1/a}_{1/b}\frac{dx}{x}\)

    44) Suppose that \(f(x)>0\) for all \(x\) and that \(f\) and \(g\) are differentiable. Use the identity \( f^g=e^{g\ln f}\) and the chain rule to find the derivative of \( f^g\).

    45) Use the previous exercise to find the antiderivative of \(h(x)=x^x(1+\ln x)\) and evaluate \(\displaystyle ∫^3_2x^x(1+\ln x)\,dx\).

    Answer:
    23

    46) Show that if \(c>0\), then the integral of \(\frac{1}{x}\) from \(ac\) to \(bc\)  \((\text{for}\,0<a<b)\) is the same as the integral of \(\frac{1}{x}\) from \(a\) to \(b\).

     

    The following exercises are intended to derive the fundamental properties of the natural log starting from the definition \(\displaystyle \ln(x)=∫^x_1\frac{dt}{t}\), using properties of the definite integral and making no further assumptions.

    47) Use the identity \(\displaystyle \ln(x)=∫^x_1\frac{dt}{t}\) to derive the identity \(\ln(\dfrac{1}{x})=−\ln x\).

    Answer:
    We may assume that \(x>1\),so \(\dfrac{1}{x}<1.\) Then, \(\displaystyle ∫^{1/x}_{1}\frac{dt}{t}\). Now make the substitution \(u=\dfrac{1}{t}\), so \(du=−\dfrac{dt}{t^2}\) and \(\dfrac{du}{u}=−\dfrac{dt}{t}\), and change endpoints: \(\displaystyle ∫^{1/x}_1\frac{dt}{t}=−∫^x_1\frac{du}{u}=−\ln x.\)

    48) Use a change of variable in the integral \(\displaystyle ∫^{xy}_1\frac{1}{t}\,dt\) to show that \(\ln xy=\ln x+\ln y\) for \( x,y>0\).

    49) Use the identity \(\displaystyle \ln x=∫^x_1\frac{dt}{x}\) to show that \(\ln(x)\) is an increasing function of \(x\) on \([0,∞)\), and use the previous exercises to show that the range of \(\ln(x)\) is \((−∞,∞)\). Without any further assumptions, conclude that \(\ln(x)\) has an inverse function defined on \( (−∞,∞).\)

    50) Pretend, for the moment, that we do not know that \(e^x\) is the inverse function of \(\ln(x)\), but keep in mind that \(\ln(x)\) has an inverse function defined on \( (−∞,∞)\). Call it \(E\). Use the identity \(\ln xy=\ln x+\ln y\) to deduce that \(E(a+b)=E(a)E(b)\) for any real numbers \(a\), \(b\).

    51) Pretend, for the moment, that we do not know that \( e^x\) is the inverse function of \(\ln x\), but keep in mind that \( \ln x\) has an inverse function defined on \((−∞,∞)\). Call it \(E\). Show that \(E'(t)=E(t).\)

    Answer:
    \(x=E(\ln(x)).\) Then, \(1=\dfrac{E'(\ln x)}{x}\) or \(x=E'(\ln x)\). Since any number \(t\) can be written \(t=\ln x\) for some \(x\), and for such \(t\) we have \(x=E(t)\), it follows that for any \(t,\,E'(t)=E(t).\)

    52) The sine integral, defined as \(\displaystyle S(x)=∫^x_0\frac{\sin t}{t}\,dt\) is an important quantity in engineering. Although it does not have a simple closed formula, it is possible to estimate its behavior for large \(x\). Show that for \(k≥1,\quad |S(2πk)−S(2π(k+1))|≤\dfrac{1}{k(2k+1)π}.\)    (Hint:   \( \sin(t+π)=−\sin t\))

    53) [T] The normal distribution in probability is given by \(p(x)=\dfrac{1}{σ\sqrt{2π}}e^{−(x−μ)^2/2σ^2}\), where \(σ\) is the standard deviation and \(μ\) is the average. The standard normal distribution in probability, \(p_s\), corresponds to \( μ=0\) and \(σ=1\). Compute the left endpoint estimates \(R_{10}\) and \(R_{100}\) of \(\displaystyle ∫^1_{−1}\frac{1}{\sqrt{2π}}e^{−x^{2/2}}\,dx.\)

    Answer:
    \(R_{10}=0.6811,\quad R_{100}=0.6827\)

     

    54) [T] Compute the right endpoint estimates \(R_{50}\) and \(R_{100}\) of \(\displaystyle ∫^5_{−3}\frac{1}{2\sqrt{2π}}e^{−(x−1)^2/8}\).

    Contributors

    • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

    • Paul Seeburger (Monroe Community College) added problems #24 - 28 in 5.5 and #28 - 29, and 31 in 5.6.