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The Fundamental Theorem of Calculus.
The two main concepts of calculus are integration and differentiation. The Fundamental Theorem of Calculus (FTC) says that these two concepts are es- sentially inverse to one another.
The fundamental theorem states that if $F$ has a continuous derivative on an interval $[a,\ b]$, then
$$
\int_{a}^{b}F'(t)dt=F(b)-F(a)\ .
$$
This form allows one to compute integrals by finding anti-derivatives.
The FTC says that integration undoes differentiation (up to a constant which is irrevocably lost when taking derivatives), in the sense that
$$
F(x)=\int_{a}^{x}\frac{d}{dt}(F(t))dt+C
$$
where $C=F(a)$ .
The second part of the fundamental theorem says that differentiation undoes integration, in the sense that
$$
f(x)=\frac{d}{dx}\int_{a}^{x}f(t)dt,
$$
where $f$ is a continuous function on an open interval containing $a$ and $x.$
Problems
1. Let $f(x)=\displaystyle \frac{1}{1+x^{4}}+a$, and let $F$ be an antiderivative of $f$, so that $F'=f.$ Find $a$ so that $F$ has exactly one critical point.
{\it Solution}. $a=-1$. Clemson Calculus Competition. $\square $
2. Let
$$
f(x)=\int_{x}^{2}\frac{1}{\sqrt{1+t^{3}}}dt.
$$
Find
$$
\int_{0}^{2}xf(x)dx.
$$
{\it Solution}. Integrate by parts to get
$$
\int_{0}^{2}xf(x)dx=-\int_{0}^{2}\frac{1}{2}x^{2}f'(x)dx=\int_{0}^{2}\frac{x^{2}}{2\sqrt{1+x^{3}}}dx.
$$
The rest is a straightforward integral after substituting for $x^{3}$. The solu- tion is $\displaystyle \frac{2}{3}. \square $
3. What function is defined by the equation
\begin{center}
$f(x)=\displaystyle \int_{0}^{x}f(t)dt+1$?
\end{center}
{\it Solution}. Larson 6.9.7 $\square $
4. Let $f$ be such that
$$
x\sin(\pi x)=\int_{0}^{x^{2}}f(t)dt.
$$
Find $f(4)$ .
{\it Solution}. Put $F(x)=\displaystyle \int_{0}^{x}f(t)dt$. Then $F(x^{2})=x\sin(\pi x)$ . Take the derivative of both sides to get $2xf(x)=\sin(\pi x)+\pi x\cos(\pi x)$ and plug in $x=2$ to find $f(4)=\pi/2$. Kenneth Roblee, Troy U. $\square $
5. If $a, b, c, d$ are polynomials, show that
$$
\int_{1}^{x}a(x)c(x)dx\int_{1}^{x}b(x)d(x)dx-\int_{1}^{x}a(x)d(x)dx\int_{1}^{x}b(x)c(x)dx
$$
is divisible by $(x-1)^{4}.$
{\it Solution}. Denote the expression in question by $F(x)$ . $F$ is a polynomial. We know $(x-1)^{4}$ divides $F$ if and only if $F'''(1)=0$. Check this by differentiation. (Larson 6.9.3). $\square $
6. Suppose that $f$ is differentiable, and that $f(x)$ is strictly increasing for $x\geq 0$. If $f(\mathrm{O})=0$, prove that $f(x)/x$ is strictly increasing for $x>0.$
{\it Solution}. Larson 6.9.12 $\square $
7. (MCMC 2005 II.5) Suppose that $ f:[0,\ \infty$) $\rightarrow[0,\ \infty$) is a differentiable function with the property that the area under the curve $y=f(x)$ from $x=a$ to $x=b$ is equal to the arclength of the curve $y=f(x)$ from $x=a$ to $x=b$. Given that $f(\mathrm{O})=5/4$, and that $f(x)$ has a minimum value on the interval $(0,\ \infty)$ , find that minimum value.
{\it Solution}. The area under the curve $y=f(x)$ from $x=a$ to $x=b$ is
$$
\int_{a}^{b}f(t)dt,
$$
and the arclength of the curve $y=f(x)$ from $x=a$ to $x=b$ is
$$
\int_{a}^{b}\sqrt{1+(f'(t))^{2}}dt.
$$
Therefore,
$$
\int_{a}^{b}f(t)dt=\int_{a}^{b}\sqrt{1+(f'(t))^{2}}dt
$$
for all nonnegative $a$ and $b$. In particular, we can write
$$
\int_{0}^{x}f(t)dt=\int_{0}^{x}\sqrt{1+(f'(t))^{2}}dt
$$
for all nonnegative $x$. Both sides of the above equation define a function of $x$, and since they are equal, their derivatives are equal; their derivatives are given by the Second Fundamental Theorem of Calculus:
$$
\frac{d}{dx}(\int_{0}^{x}f(t)dt)=\frac{d}{dx}(\int_{0}^{x}\sqrt{1+(f'(t))^{2}}dt)\ ,
$$
i. e.,
$$
f(x)=\sqrt{1+(f'(x))^{2}}.
$$
So, we are looking for a function $y$ which satisifies the differential equation
$$
y=\sqrt{1+(y')^{2}}.
$$
This equation is separable:
\begin{center}
$y=\sqrt{1+(y')^{2}}\Rightarrow y^{2}=1+(y')^{2}$ (1)
$\Rightarrow(y')^{2}=y^{2}-1$ (2)
$\Rightarrow y'=\sqrt{y^{2}-1}$ (3)
$\displaystyle \Rightarrow\frac{dy}{\sqrt{y^{2}-1}}=dx$. (4)
\end{center}
Integrating both sides yields
$$
\int\frac{dy}{\sqrt{y^{2}-1}}=\int dx\Rightarrow\ln|y+\sqrt{y^{2}-1}|=x+C
$$
(where the first integral is evaluated using the trig substitution $ y=\sec\theta$ and the two arbitrary constants of integration are combined into one con- stant on the right hand side). Next, since $f(\mathrm{O})=5/4$ is positive, we can drop the absolute value, and solve for $y$:
$\ln(y+\sqrt{y^{2}-1})=x+C\Rightarrow y+\sqrt{y^{2}-1}=e^{x+C}=Ae^{x}$ (where $A=e^{C}$)
$$
\Rightarrow\sqrt{y^{2}-1}=Ae^{x}-y
$$
(5)
$$
\Rightarrow y^{2}-1=(Ae^{x}-y)^{2}=A^{2}e^{2x}-2Aye^{x}+\mathrm{y}6)
$$
$$
\Rightarrow-1=A^{2}e^{2x}-2Aye^{x}
$$
(7)
\begin{center}
$\Rightarrow 2Aye^{x}=A^{2}e^{2x}+1$ (8)
$\displaystyle \Rightarrow y=\frac{A^{2}e^{2x}+1}{2Ae^{x}}=\frac{A}{2}e^{x}+\frac{1}{2A}e^{-x}$. (9)
\end{center}
Using $f(\mathrm{O})=5/4$, we find
$\displaystyle \frac{5}{4}=\frac{A}{2}+\frac{1}{2A}\Rightarrow A=\frac{1}{2}$ or 2.
This gives two possible functions:
\begin{center}
$y=\displaystyle \frac{1}{4}e^{x}+e^{-x}$ or $y=e^{x}+\displaystyle \frac{1}{4}e^{-x}.$
\end{center}
This latter has a minimum at $x=-\ln 2$, which is not positive, so we reject that function. The former has a minimum at $x=\ln 2$, and the $y$ value is 1. Note: One could also deduce from the differential equation $y'=\sqrt{y^{2}-1}$ that at the minimum value, since $y'=0$, the $y$-value must be 1. $\square $
8. (MCMC 2006 I.5) Let $f(t)$ and $f(t)$ be differentiable on $[a,\ x]$ and for each $x$ suppose there is a number $c_{x}$ such that $a