미적분의 기본정리

By the Fundamental Theorem of Calculus Part 1,

$$f'(x) = 100(x^2-3x+2)e^{-x^2} = 100(x-1)(x-2)e^{-x^2}$$

As $f(x)$ is increasing whenever $f'(x)>0$ and $100 e^{-x^2}$ is always strictly bigger than $0$, we have $f(x)$ increasing if and only if $(x-1)(x-2)>0$, which is the case if and only if $(x-1)$ and $(x-2)$ are of the same sign. Both are positive when $x>2$ and both are negative when $x<1$. So $f(x)$ is increasing when $-\infty<x<1$ and when $2<x<\infty$.

Remark: even without the Fundamental Theorem of Calculus, since $f(x)$ is the area under a curve from 1 to $x$, $f(x)$ is increasing when the curve is above the $x$-axis (because we're adding positive area), and it's decreasing when the curve is below the $x$-axis (because we're adding negative area).

Applying $\frac{\mathrm{d}}{\mathrm{d}x}$ to both sides of $x^2=1+\int_1^x f(t)\ \mathrm{d}t$ gives, by the Fundamental Theorem of Calculus Part 1, $2x=f(x)$.

Apply $\frac{\mathrm{d}}{\mathrm{d}x}$ to both sides of $x \sin(\pi x) = \int_0^x f(t)\, \mathrm{d}t$. Then, by the Fundamental Theorem of Calculus Part 1,

$$\begin{aligned} &&f(x) =\frac{\mathrm{d}}{\mathrm{d}x} \int_0^x f(t)\, \mathrm{d}t&=\frac{\mathrm{d}}{\mathrm{d}x}\big\{x \sin(\pi x)\big\} \\ \implies&& f(x) &= \frac{\mathrm{d}}{\mathrm{d}x}\big\{x \sin(\pi x)\big\} =\sin(\pi x)+\pi x\cos(\pi x) \\ \implies&& f(4)&= \sin(4\pi)+4\pi\cos(4\pi)=4\pi \end{aligned}$$

Recall that “$+C$" means that we can add any constant to the function. Since $\tan^2 x = \sec^2 x - 1$, Students A and B have equivalent answers: they only differ by a constant.

So, if one is right, both are right; if one is wrong, both are wrong. We check Student A's work: $$\frac{\mathrm{d}}{\mathrm{d}x}\{\tan^2 x + x + C\}=\frac{\mathrm{d}}{\mathrm{d}x}\{\tan^2 x\} +1 + 0 = f(x)-1+1=f(x)$$ So, Student A's answer is indeed an anditerivative of $f(x)$. Therefore, both students ended up with the correct answer.

Remark: it is a frequent occurrence that equivalent answers might look quite different. As you are comparing your work to others', this is a good thing to keep in mind!

(a) When $x=3$,

$$\begin{aligned} F(3)&=\displaystyle\int_0^3 3^3 \sin(t)\ \mathrm{d}t=27\int_0^3 \sin t \ \mathrm{d}t \\ \text{Using the Fundamental Theorem of Calculus Part 2,} \\ &=27\left[-\cos t\right]_{t=0}^{t=3} =27\left[-\cos 3 - (-\cos 0)\right]\\ &=27(1-\cos 3 ) \end{aligned}$$ (b) Since the integration is with respect to $t$, the $x^3$ term can be moved outside the integral. That is: for the purposes of the integral, $x^3$ is a constant (although for the purposes of the derivative, it certainly is not).

$$\begin{aligned} F(x)&=\displaystyle\int_0^x x^3 \sin(t)\ \mathrm{d}t = x^3 \int_0^x \sin(t)\ \mathrm{d}t \\ \text{Using the product rule and the Fundamental Theorem of Calculus Part 1,} \\ F'(x)&=x^3\cdot \sin(x) + 3x^2 \int_0^x \sin(t)\ \mathrm{d}t\\ &=x^3\sin(x)+3x^2\left[-\cos(t)\right]_{t=0}^{t=x}\\ &=x^3\sin(x)+3x^2[-\cos(x)-(-\cos(0))]\\ &=x^3\sin (x) + 3x^2[1-\cos (x)] \end{aligned}$$

Remark: Since $x$ and $t$ play different roles in our problem, it's crucial that they have different names. This is one reason why we should avoid the common mistake of writing $\int_a^x f(x)\mathrm{d}x$ when we mean $\int_a^x f(t)\mathrm{d}t$.

(a) The net signed area enclosed by $f$ and the $x$-axis on the interval $[2,5]$ is $\int_2^5 f(x) \, dx$. The figure below shows a graph of $f$ on the interval $[0,5]$. To calculate the definite integral of $f$ on the interval $[2,5]$, we consider the integrals on the subintervals $[2,3]$ and $[3,5]$. Doing so and using the Fundamental Theorem of Calculus, $$\begin{aligned}\int_2^5 f(x) \, dx &= \int_2^3 f(x) \, dx + \int_3^5 f(x) \, dx \\ &= \int_2^3 -x+3 \, dx + \int_3^5 x^2-8x+15 \, dx \\ &= \left. \left( -\frac{1}{2}x^2 + 3x \right)\right|_2^3 + \left. \left( \frac{1}{3}x^3 - 4x^2 + 15x \right)\right|_3^5 \\ &= \left(\frac{9}{2}-4\right) + \left( \frac{50}{3}-18 \right) \\ &= \frac{127}{6} - 22 = -\frac{5}{6}\end{aligned}\,.$$ (b) To find the average value of $f$ on $[0,5]$ we need to first calculate $$\int_0^5 f(x) \, dx\,.$$ Using our work in (a) and considering the subintervals $[0,2]$ and $[2,5]$, $$\begin{aligned}\int_0^5 f(x) \, dx &= \int_0^2 f(x) \, dx + \int_2^5 f(x) \, dx \\ &= \int_0^2 -x^2 + 2x + 1 \, dx - \frac{5}{6} \\ &= \left. \left( -\frac{1}{3}x^3 + x^2 + x \right)\right|_0^2 - \frac{5}{6} \\ &= \frac{10}{3} - \frac{5}{6} \\ &= \frac{5}{2}\end{aligned}\,.$$ Therefore, the average value of $f$ on $[0,5]$ is $$\displaystyle f_{\operatorname{AVG} [0,5]} = \frac{1}{5-0} \int_0^5 f(x) \, dx = \left(\frac{1}{5}\right) \left(\frac{5}{2}\right) = \frac{1}{2}\,.$$ (c) Since $\int_0^5 f(x) \, dx = \frac{5}{2}$, we need a function $g$ that matches $f$ but that is also defined on $[5,7]$ so that $\int_5^7 g(x) \, dx = -\frac{5}{2}$. To preserve continuity, we will choose a formula for $g$ that also satisfies $g(5) = 0$. It's natural to pick a linear function whose graph encloses $5/2$ units of area, so one that has a slope of $5/4$. To check, we compute $$\begin{aligned}\int_5^7 -\frac{5}{4}(x - 5) \, dx &= -\left(\frac{5}{4}\right) \left. \left(\frac{x^2}{2} - 5x\right) \right|_5^7 \\ &= -\left(\frac{5}{4}\right) \left( \frac{49-25}{2} - 5(7-5) \right) \\ &= -\left(\frac{5}{4}\right) \left( 12 - 10 \right) \\ &= -\left(\frac{5}{4}\right) \left( 2 \right) = -\frac{5}{2}\end{aligned}\,.$$ So $g$ defined so that $g(x) = f(x)$ for $0 \le x < 5$ and $g(x) = -\frac{5}{4}(x - 5)$ on $5 \le x \le 7$ has the desired property. A graph of $g$ is shown in the figure below.