Calc-OkQuestion
Most-appropriate topic codes (CED):
• TOPIC 8.1: Average Value of a Function — part (b)
• TOPIC 5.2 / 5.5: EVT, Critical Points & Candidates Test — part (c)
• TOPIC 4.1: Interpreting the Meaning of the Derivative in Context — part (d)
▶️ Answer/Explanation
\[ \int_{0}^{5}\!\!E(t)\,dt=\int_{0}^{5}\!\!\Big(20+15\sin\!\tfrac{\pi t}{6}\Big)dt =\underbrace{\int_{0}^{5}\!20\,dt}_{=100}+\underbrace{15\int_{0}^{5}\!\sin\!\tfrac{\pi t}{6}\,dt}_{(*)}. \]
For \((*)\): \(\displaystyle \int \sin(kt)\,dt=-\frac{1}{k}\cos(kt)\), with \(k=\tfrac{\pi}{6}\).
\[ 15\left[-\frac{1}{k}\cos(kt)\right]_{0}^{5} =15\left[-\frac{6}{\pi}\cos\!\left(\tfrac{\pi t}{6}\right)\right]_{0}^{5} =-\frac{90}{\pi}\Big(\cos\tfrac{5\pi}{6}-\cos 0\Big). \]
\(\cos\tfrac{5\pi}{6}=-\tfrac{\sqrt3}{2}\), \(\cos 0=1\) \(\Rightarrow\) term \(=\dfrac{90}{\pi}\!\left(1+\tfrac{\sqrt3}{2}\right)\).
Total: \[ 100+\frac{90}{\pi}\!\left(1+\frac{\sqrt3}{2}\right)\approx 153.457690\ \text{fish}. \]
Nearest whole number: 153 fish. (Applied accumulation context: Topic 8.3.)
(b) Average number leaving on \([0,5]\)
Average value on \([a,b]\): \(\displaystyle \frac{1}{b-a}\int_{a}^{b} f(x)\,dx\). (Topic 8.1.)
\[ \text{Average}=\frac{1}{5}\int_{0}^{5}\!\!L(t)\,dt =\frac{1}{5}\!\left(\int_{0}^{5}\!4\,dt+\int_{0}^{5}\!2^{\,0.1t^{2}}\,dt\right) =\frac{1}{5}\!\left(20+\int_{0}^{5}\!2^{\,0.1t^{2}}\,dt\right). \]
The integral \(\displaystyle \int_{0}^{5}\!2^{\,0.1t^{2}}\,dt\) has no elementary antiderivative; using numerical integration (e.g., refined Riemann/Simpson) gives \(\displaystyle \int_{0}^{5}\!2^{\,0.1t^{2}}\,dt\approx 10.29519\).
Hence \[ \text{Average}\approx \frac{20+10.29519}{5}=6.059038\ \text{fish/hour}. \]
(c) Time when the number of fish is greatest on \([0,8]\)
Let \(N(t)\) be the number of fish in the lake since midnight. Then \(N'(t)=E(t)-L(t)\).
A maximum of \(N(t)\) occurs where \(N'(t)=0\) and changes from \(+\) to \(-\) (EVT/critical points + candidates test). (Topics 5.2 & 5.5.)
Solve \(E(t)-L(t)=0\): \(20+15\sin\!\big(\tfrac{\pi t}{6}\big)=4+2^{\,0.1t^{2}}\).
Using Newton’s method starting near \(t_0=6.2\):
• \(f(t)=E(t)-L(t)\), \(f'(t)=E'(t)-L'(t)\), with \(E'(t)=\tfrac{15\pi}{6}\cos\!\big(\tfrac{\pi t}{6}\big)\), \(L'(t)=2^{\,0.1t^{2}}\ln 2\cdot(0.2t)\).
• \(t_{1}=t_{0}-\dfrac{f(t_{0})}{f'(t_{0})}\approx 6.2037\).
Refined value: \(t\approx \mathbf{6.20356}\) hours. Sign check: for \(t<6.20356\), \(E-L>0\); for \(t>6.20356\), \(E-L<0\).
Therefore the greatest number of fish occurs at \(t\approx 6.204\) h.
(d) Is the rate \(N'(t)\) increasing or decreasing at \(t=5\)?
\(N”(t)=E'(t)-L'(t)\). Compute at \(t=5\):
\(E'(t)=\dfrac{15\pi}{6}\cos\!\big(\tfrac{\pi t}{6}\big)=\dfrac{5\pi}{2}\cos\!\big(\tfrac{\pi t}{6}\big)\).
\(\cos\!\big(\tfrac{5\pi}{6}\big)=-\tfrac{\sqrt3}{2}\Rightarrow E'(5)=-\dfrac{5\pi\sqrt3}{4}\approx -6.8017.\)
\(L'(t)=2^{\,0.1t^{2}}\ln 2\cdot(0.2t)\Rightarrow L'(5)=2^{2.5}\ln 2\cdot 1= (4\sqrt2)(\ln 2)\approx 5.656854\times 0.693147\approx 3.9209.\)
\(N”(5)=E'(5)-L'(5)\approx -6.8017-3.9209=-10.7226<0\).
Therefore the rate \(N'(t)\) is decreasing at \(t=5\) A.M.
Calc-Ok Question
Let \(S\) be the region bounded by the polar curve \(r(\theta)=3\sqrt{\theta}\,\sin(\theta^{2})\) for \(0\le \theta \le \sqrt{\pi}\), as shown.
Most-appropriate topic codes (CED):
• TOPIC 9.9: Area Bounded by Two Polar Curves (BC) — part (d)
• TOPIC 8.1: Average Value of a Function — part (b)
▶️ Answer/Explanation
Polar-area formula (single curve): \(\displaystyle \text{Area}=\frac12\int_{\alpha}^{\beta}\!\big(r(\theta)\big)^{2}\,d\theta\).
\[ A=\frac12\int_{0}^{\sqrt{\pi}}\!\!\big(3\sqrt{\theta}\,\sin(\theta^{2})\big)^{2} d\theta =\frac12\int_{0}^{\sqrt{\pi}}\! 9\theta\,\sin^{2}(\theta^{2})\,d\theta. \]
Use \(\sin^{2}u=\tfrac12(1-\cos 2u)\) with \(u=\theta^{2}\):
\[ A=\frac12\int_{0}^{\sqrt{\pi}}\!9\theta\cdot \frac{1-\cos(2\theta^{2})}{2}\,d\theta =\frac{9}{4}\int_{0}^{\sqrt{\pi}}\!\theta\,d\theta-\frac{9}{4}\cdot\frac12\int_{0}^{\sqrt{\pi}}\!\theta\cos(2\theta^{2})\,d\theta. \]
For \(\displaystyle \int \theta\cos(2\theta^{2})\,d\theta\), let \(w=\theta^{2}\Rightarrow dw=2\theta\,d\theta\):
\(\displaystyle \int \theta\cos(2\theta^{2})\,d\theta=\tfrac12\int \cos(2w)\,dw=\tfrac14\sin(2w)=\tfrac14\sin(2\theta^{2}).\)
Therefore an antiderivative for \(9\theta\sin^{2}(\theta^{2})\) is \(\displaystyle \frac{9}{4}\theta^{2}-\frac{9}{8}\sin(2\theta^{2})\). Hence
\[ A=\frac12\Big[\frac{9}{4}\theta^{2}-\frac{9}{8}\sin\!\big(2\theta^{2}\big)\Big]_{0}^{\sqrt{\pi}} =\frac12\cdot\frac{9}{4}\pi=\boxed{\frac{9\pi}{8}}\approx \mathbf{3.534292}. \]
(b) Average distance from the origin
Average value on \([0,\sqrt{\pi}]\): \(\displaystyle \text{Avg}=\frac{1}{\sqrt{\pi}}\int_{0}^{\sqrt{\pi}}\! r(\theta)\,d\theta\). (Topic 8.1.)
\[ \text{Avg}=\frac{1}{\sqrt{\pi}}\int_{0}^{\sqrt{\pi}}3\sqrt{\theta}\,\sin(\theta^{2})\,d\theta. \]
Substitute \(x=\theta^{2}\Rightarrow dx=2\theta\,d\theta\), \(d\theta=\frac{dx}{2\sqrt{x}}\), \(\sqrt{\theta}=x^{1/4}\):
\[ \text{Avg}=\frac{1}{\sqrt{\pi}}\int_{0}^{\pi}\frac{3}{2}\,x^{-1/4}\sin x\,dx, \] which has no elementary antiderivative; evaluating numerically (e.g., Simpson’s rule / calculator) gives
\[ \boxed{\text{Avg}\approx 1.579933\ \text{fish-units}} \ (\text{about } \mathbf{1.580}). \]
(c) Line through the origin dividing \(S\) into equal areas
A line with slope \(m>0\) through the origin corresponds to the polar angle \(\theta=\arctan m\).
Equal areas \(\Rightarrow\) area from \(0\) to \(\arctan m\) equals half of the total area:
\[ \boxed{\;\frac12\int_{0}^{\arctan m}\!\!\big(r(\theta)\big)^{2}\,d\theta \;=\;\frac12\cdot\frac12\int_{0}^{\sqrt{\pi}}\!\!\big(r(\theta)\big)^{2}\,d\theta\;} \] with \(r(\theta)=3\sqrt{\theta}\,\sin(\theta^{2})\).
(d) \(\displaystyle \lim_{k\to\infty}A(k)\) for the circle \(r=k\cos\theta\)
The circle \(r=k\cos\theta\) (center \((k/2,0)\), radius \(k/2\)) expands and, as \(k\to\infty\), includes all points with \(x\ge 0\), i.e., \(-\tfrac{\pi}{2}\le \theta\le \tfrac{\pi}{2}\).
Since \(S\) uses \(0\le \theta\le \sqrt{\pi}\) and \(\tfrac{\pi}{2}<\sqrt{\pi}\), the limiting overlap is the portion of \(S\) with \(0\le \theta\le \tfrac{\pi}{2}\).
Therefore \[ \lim_{k\to\infty}A(k) =\frac12\int_{0}^{\pi/2}\!\!\big(r(\theta)\big)^{2}d\theta =\frac12\int_{0}^{\pi/2}\! 9\theta\sin^{2}(\theta^{2})\,d\theta. \] Using the antiderivative from part (a), \[ \lim_{k\to\infty}A(k) =\frac12\Big[\frac{9}{4}\theta^{2}-\frac{9}{8}\sin(2\theta^{2})\Big]_{0}^{\pi/2} =\frac{9}{8}\left(\frac{\pi^{2}}{4}\right)-\frac{9}{16}\sin\!\Big(\frac{\pi^{2}}{2}\Big). \] Numerical value: \[ \boxed{\lim_{k\to\infty}A(k)\approx \mathbf{3.324}}. \]
No-Calc Question 3
The continuous function \(f\) is defined on the closed interval \(-6\le x\le 5\). The figure shows a portion of the graph of \(f\): two line segments and a quarter of a circle centered at \((5,3)\). It is known that the point \((3,\,3-\sqrt5)\) is on the graph of \(f\).
Most-appropriate topic codes (CED):
• TOPIC 8.4: Areas via Definite Integrals (geometric reasoning with signed area) — part (a)
• TOPIC 5.5: Candidates Test for Absolute Extrema on a Closed Interval — part (c)
• TOPIC 4.1: Interpreting Derivatives / continuity & direct evaluation in limits — part (d)
▶️ Answer/Explanation
(a) Find \(\displaystyle \int_{-6}^{-2} f(x)\,dx\)
From the figure, decompose the signed area on \([-2,5]\):
• On \([-2,0]\), the segment forms a right triangle of base \(2\) and height \(2\) (below the \(x\)-axis). Signed area \(=\,-\tfrac12(2)(2)=-2\).
• On \([0,3]\), the line segment from \((0,-2)\) to \((3,3)\) gives a triangle of base \(3\) and height \(5\). Signed area \(=\tfrac12(3)(5)=\tfrac{15}{2}\).
• On \([3,5]\), the arc is a quarter circle of radius \(3\) centered at \((5,3)\); the area under the curve equals the area of the \(3\times3\) square minus the quarter-disk: \(9-\dfrac{9\pi}{4}\).
Hence \[ \int_{-2}^{5}\!f(x)\,dx=(-2)+\frac{15}{2}+\left(9-\frac{9\pi}{4}\right)=11-\frac{9\pi}{4}. \] From the prompt we have total “given value” \(7\) and split the interval \([-6,5]\) at \(-2\):
\[ 7=\int_{-6}^{-2}\!f(x)\,dx+\int_{-2}^{5}\!f(x)\,dx. \] Therefore \[ \int_{-6}^{-2}\!f(x)\,dx = 7-\left(11-\frac{9\pi}{4}\right)=\boxed{\frac{9\pi}{4}-4}. \]
(b) Evaluate \(\displaystyle \int_{3}^{5}\!\big(2f'(x)+4\big)\,dx\)
Use linearity and the Fundamental Theorem of Calculus (FTC):
\[ \int_{3}^{5}\!\big(2f'(x)+4\big)\,dx =2\!\int_{3}^{5}\!f'(x)\,dx+4(5-3) =2\,[f(x)]_{3}^{5}+8 =2\big(f(5)-f(3)\big)+8. \] From the graph: \(f(5)=0\) and \(f(3)=3-\sqrt5\). Thus \[ \boxed{2+2\sqrt5}. \]
(c) Absolute maximum of \(g(x)=\displaystyle\int_{2}^{x}\!f(t)\,dt\) on \([-2,5]\)
\(g'(x)=f(x)\). Critical points where \(f(x)=0\) occur at \(x=-1,\; x=\tfrac12,\; x=5\) (from the figure). Also check the endpoints \(x=-2\) and \(x=5\)
Accumulate signed area from \(x=2\) left/right to each candidate (or build a running table):
\[ \begin{array}{c|c} x & g(x)=\int_{2}^{x}\!f(t)\,dt\\\hline -2 & 0\\ 0 & 1\\ -1 & \tfrac12\\ \tfrac12 & \tfrac14\\ 4 & 5\\ 5 & 11-\dfrac{9\pi}{4} \end{array} \] Comparing values, the absolute maximum on \([-2,5]\) is \[ \boxed{g(5)=\,11-\dfrac{9\pi}{4}}. \]
(d) \(\displaystyle \lim_{x\to 1}\frac{10^{x}-3f'(x)}{\,f(x)-\arctan x\,}\)
Evaluate by continuity of \(10^{x}\), \(f\), \(f’\), and \(\arctan x\) at \(x=1\) (the denominator is nonzero):
From the graph, \(f(1)=1\) and the slope at \(x=1\) is \(f'(1)=2\). Then \[ \lim_{x\to 1}\frac{10^{x}-3f'(x)}{\,f(x)-\arctan x\,} =\frac{10^{1}-3\cdot 2}{\,1-\arctan(1)\,} =\boxed{\frac{4}{\,1-\tfrac{\pi}{4}\,}}. \]
No-Calc Question 4
A cylindrical barrel with a diameter of \(2\) feet contains collected rainwater, as shown in the figure above. The water drains out through a valve (not shown) at the bottom of the barrel. The rate of change of the height \(h\) of the water in the barrel with respect to time \(t\) is modeled by \[ \frac{dh}{dt}=-\frac{1}{10}\sqrt{h}, \] where \(h\) is measured in feet and \(t\) is measured in seconds. (The volume \(V\) of a cylinder with radius \(r\) and height \(h\) is \(V=\pi r^{2}h\).)
Most-appropriate topic codes (CED):
• TOPIC 7.1: Modeling Situations with Differential Equations — setup/interpretation of \( \tfrac{dh}{dt} \)
• TOPIC 7.6–7.7: Separation of Variables; Particular Solutions with Initial Conditions — solve for \(h(t)\)
▶️ Answer/Explanation
Diameter \(=2\ \Rightarrow\) radius \(r=1\ \text{ft}\).
\(V=\pi r^{2}h=\pi(1)^{2}h=\pi h\).
Differentiate with respect to \(t\): \(\displaystyle \frac{dV}{dt}=\pi\frac{dh}{dt}\).
At \(h=4\): \(\displaystyle \frac{dh}{dt}=-\frac{1}{10}\sqrt{4}=-\frac{1}{10}\cdot 2=-\frac{1}{5}\ \text{ft/s}\).
Therefore \[ \boxed{\frac{dV}{dt}=\pi\!\left(-\frac{1}{5}\right)=-\frac{\pi}{5}\ \text{ft}^3/\text{s}}. \]
(b) Is \(\dfrac{dh}{dt}\) increasing or decreasing at \(h=3\)?
Compute the second derivative with respect to \(t\):
\(\displaystyle \frac{dh}{dt}=-\frac{1}{10}h^{1/2}\).
Differentiate via chain rule: \(\displaystyle \frac{d^{2}h}{dt^{2}}=\frac{d}{dt}\!\left(-\frac{1}{10}h^{1/2}\right)=-\frac{1}{20}h^{-1/2}\cdot\frac{dh}{dt}\).
Substitute \(\displaystyle \frac{dh}{dt}=-\frac{1}{10}\sqrt{h}\):
\[ \frac{d^{2}h}{dt^{2}}=-\frac{1}{20}h^{-1/2}\left(-\frac{1}{10}\sqrt{h}\right)=\frac{1}{200}>0. \] Since \( \dfrac{d^{2}h}{dt^{2}}>0\) (for \(h>0\)), the rate \( \dfrac{dh}{dt}\) is increasing when \(h=3\ \text{ft}\).
(c) Solve for \(h(t)\) using separation, given \(h(0)=5\)
Start with \(\displaystyle \frac{dh}{dt}=-\frac{1}{10}\sqrt{h}\).
Separate variables: \(\displaystyle \frac{dh}{\sqrt{h}}=-\frac{1}{10}\,dt\).
Integrate both sides: \[ \int \! h^{-1/2}\,dh=\int\! -\frac{1}{10}\,dt \quad\Rightarrow\quad 2\sqrt{h}=-\frac{1}{10}t+C. \] Apply the initial condition \(h(0)=5\): \(2\sqrt{5}=C\).
Thus \(2\sqrt{h}=-\dfrac{1}{10}t+2\sqrt{5}\ \Rightarrow\ \sqrt{h}=-\dfrac{t}{20}+\sqrt{5}\).
Square to solve for \(h\): \[ \boxed{\,h(t)=\left(-\frac{t}{20}+\sqrt{5}\right)^{2}\,}. \] Domain note: this model is valid while \(h\ge 0\), i.e., for \(0\le t\le 20\sqrt{5}\).
No-Calc Question 5
Most-appropriate topic codes (CED):
• TOPIC 3.1: The Chain Rule — derivative of a reciprocal/composite (support for part (a)).
• TOPIC 6.12: Integrating Using Linear Partial Fractions (BC) — decomposition/integration (part (b)).
• TOPIC 6.13: Evaluating Improper Integrals (BC) — limit definition of convergence/divergence (part (c)).
▶️ Answer/Explanation
(a) Slope \(=f'(0)=6\) ⇒ solve for \(k\) (\(k>0\))
\(f(x)=(x^{2}-2x+k)^{-1}\). By chain/power rule, \[ f'(x)=-1\cdot (x^{2}-2x+k)^{-2}\cdot(2x-2)=\frac{-(2x-2)}{(x^{2}-2x+k)^{2}}. \] Evaluate at \(x=0\): \[ f'(0)=\frac{-(-2)}{k^{2}}=\frac{2}{k^{2}}. \] Set \(f'(0)=6\Rightarrow \dfrac{2}{k^{2}}=6\Rightarrow k^{2}=\dfrac{1}{3}\Rightarrow \boxed{k=\dfrac{1}{\sqrt{3}}}\ (\text{since }k>0). \)
(b) \(\displaystyle \int_{0}^{1} f(x)\,dx\) for \(k=-8\)
Then \(f(x)=\dfrac{1}{x^{2}-2x-8}=\dfrac{1}{(x-4)(x+2)}\). Use linear partial fractions (BC): find \(A,B\) with \[ \frac{1}{(x-4)(x+2)}=\frac{A}{x-4}+\frac{B}{x+2} \ \Rightarrow\ 1=A(x+2)+B(x-4). \] Solve: set \(x=4\Rightarrow 1=6A\Rightarrow A=\dfrac{1}{6}\); set \(x=-2\Rightarrow 1=-6B\Rightarrow B=-\dfrac{1}{6}\). Hence \[ \int_{0}^{1}\!f(x)\,dx =\int_{0}^{1}\!\left(\frac{1}{6}\frac{1}{x-4}-\frac{1}{6}\frac{1}{x+2}\right)\!dx =\left[\frac{1}{6}\ln|x-4|-\frac{1}{6}\ln|x+2|\right]_{0}^{1}. \] Evaluate: \[ =\left(\tfrac{1}{6}\ln 3-\tfrac{1}{6}\ln 3\right)-\left(\tfrac{1}{6}\ln 4-\tfrac{1}{6}\ln 2\right) =\boxed{\tfrac{1}{6}\ln 2}. \]
(c) \(\displaystyle \int_{0}^{2} f(x)\,dx\) for \(k=1\) (improper)
Now \(f(x)=\dfrac{1}{x^{2}-2x+1}=\dfrac{1}{(x-1)^{2}}\), which is unbounded at \(x=1\). Split and evaluate as limits: \[ \int_{0}^{2}\frac{1}{(x-1)^{2}}\,dx =\lim_{b\to 1^{-}}\int_{0}^{b}\frac{1}{(x-1)^{2}}\,dx +\lim_{b\to 1^{+}}\int_{b}^{2}\frac{1}{(x-1)^{2}}\,dx. \] Antiderivative: \(\displaystyle \int \frac{1}{(x-1)^{2}}dx=-\,\frac{1}{x-1}+C\). Thus \[ \lim_{b\to 1^{-}}\!\Big[-\tfrac{1}{x-1}\Big]_{0}^{b} =\lim_{b\to 1^{-}}\!\left(-\frac{1}{b-1}+1\right)=+\infty, \quad \lim_{b\to 1^{+}}\!\Big[-\tfrac{1}{x-1}\Big]_{b}^{2} =\lim_{b\to 1^{+}}\!\left(1+\frac{1}{b-1}\right)=+\infty. \] Since at least one (indeed both) limits diverge, the integral \[ \boxed{\text{diverges (does not exist as a finite value).}} \]
No-Calc Question 6
A function \(f\) has derivatives of all orders for all real numbers \(x\). A portion of the graph of \(f\) is shown above, along with the line tangent to the graph of \(f\) at \(x=0\). Selected derivatives of \(f\) at \(x=0\) are given in the table below.
| \(n\) | \(f^{(n)}(0)\) |
|---|---|
| 2 | \(3\) |
| 3 | \(-\dfrac{23}{2}\) |
| 4 | \(54\) |
Most-appropriate topic codes (CED):
• TOPIC 10.14: Finding Taylor or Maclaurin Series for a Function; using known series like \(e^{x}\) — part (b).
• TOPIC 10.15: Representing Functions as Power Series (operations with series/products) — part (b).
• TOPIC 10.10: Alternating Series Error Bound — part (d).
▶️ Answer/Explanation
Information from the figure & tangent at \(x=0\)
From the graph: \(f(0)=3\). From the tangent line at \(x=0\): slope \(f'(0)=-2\).
(a) Third-degree Taylor polynomial for \(f\) at \(x=0\)
General form: \(T_{3}(x)=f(0)+f'(0)x+\dfrac{f”(0)}{2!}x^{2}+\dfrac{f^{(3)}(0)}{3!}x^{3}\). (Taylor approximation: Topic 10.11.)
Substitute \(f(0)=3\), \(f'(0)=-2\), \(f”(0)=3\), \(f^{(3)}(0)=-\dfrac{23}{2}\):
\[ \boxed{\,T_{3}(x)=3-2x+\frac{3}{2}x^{2}-\frac{23}{12}x^{3}\, }. \]
(b) Series for \(e^{x}\) and the 2nd-degree Taylor polynomial for \(e^{x}f(x)\)
Maclaurin for \(e^{x}\): \(1+x+\dfrac{x^{2}}{2!}+\cdots\). (Known series foundation: Topic 10.14.)
Multiply \(e^{x}\) (through \(x^{2}\)) by \(T_{3}\) (through \(x^{2}\)) and keep terms \(\le x^{2}\):
\[ (1+x+\tfrac{x^{2}}{2})\big(3-2x+\tfrac{3}{2}x^{2}\big) =3\,(1+x+\tfrac{x^{2}}{2})-2x(1+x)+\tfrac{3}{2}x^{2}(1). \] Collect coefficients:
constant \(=3\); \(x\)-term \(=3x-2x=x\); \(x^{2}\)-term \(=\tfrac{3}{2}-2+\tfrac{3}{2}=1\).
Therefore the second-degree Taylor polynomial is \[ \boxed{\,3+x+x^{2}\, }. \] (Operations with series/products: Topic 10.15.)
(c) Approximate \(h(1)\) with \(T_{3}\)
\(h(x)=\displaystyle\int_{0}^{x}f(t)\,dt \approx \int_{0}^{x}T_{3}(t)\,dt.\)
\[ h(1)\approx \int_{0}^{1}\!\Big(3-2t+\tfrac{3}{2}t^{2}-\tfrac{23}{12}t^{3}\Big)\,dt =\Big[3t-t^{2}+\tfrac{1}{2}t^{3}-\tfrac{23}{48}t^{4}\Big]_{0}^{1} =\boxed{\frac{97}{48}}\ \ (\approx 2.020833). \]
(d) Alternating series error bound for \(h(1)\)
The Maclaurin series for \(f\) has next term \(\dfrac{f^{(4)}(0)}{4!}x^{4}=\dfrac{54}{24}x^{4}=\dfrac{9}{4}x^{4}\) (positive). Integrating term-by-term to get the series for \(h(1)\), the first omitted term’s magnitude is \[ \left|\int_{0}^{1}\frac{9}{4}t^{4}\,dt\right|=\frac{9}{4}\cdot\frac{1}{5}=\boxed{\frac{9}{20}}=0.45. \] Because the terms in the series for \(h(1)\) alternate and decrease to \(0\), the alternating series error bound (Topic 10.10) guarantees \[ \big|\,h(1)-\tfrac{97}{48}\,\big|\le \frac{9}{20}\le 0.45. \]