적분 전체 문제

$9/2$

$343/6$

Notice $f(x)$ is nonnegative for an interval covering the left part of the graph, and negative on the right part of the graph. That means its antiderivative is increasing for the left interval, then decreasing for the right interval. This applies to $A(x)$ and $C(x)$, but not $B(x)$.

There are only three points where $A(x)$ has a horizontal tangent line: at its global maximum and the endpoints of the interval shown. By contrast, $C(x)$ has a horizontal tangent line in four places: at its global maximum, at its inflection point, and at the endpoints of the interval shown. Since $f(x)=0$ four times (and these line up with the horizontal portions of $C(x)$) we conclude $C(x)$ is the antiderivative of $f(x)$.

Let's recall the definitions of even and odd functions: $f(x)$ is even if $f(-x)=f(x)$ for every $x$ in its domain, and $f(x)$ is odd if $f(-x)=-f(x)$ for every $x$ in its domain.

Let $h(x)=f(x)\cdot g(x)$.

• **even $\times$ even: ** If $f$ and $g$ are both even, then $h(-x)=f(-x)\cdot g(-x) = f(x)\cdot g(x)=h(x)$, so their product is even.
• **odd $\times$ odd: ** If $f$ and $g$ are both odd, then $h(-x)=f(-x)\cdot g(-x) =[- f(x)]\cdot [-g(x)]=f(x)\cdot g(x)=h(x)$, so their product is even.
• **even $\times$ odd: ** If $f$ is even and $g$ is odd, then $h(-x)=f(-x)\cdot g(-x) = f(x)\cdot[- g(x)]=-[f(x)\cdot g(x)]=-h(x)$, so their product is odd. Because multiplication is commutative, the order we multiply the functions in doesn't matter.

We note that the table would be the same as if we were adding (not multiplying) even and odd numbers (not functions).

(a) Using the product rule, followed by the chain rule, followed by the Fundamental Theorem of Calculus Part 1,

$$\begin{aligned} f'(x) & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \mathrm{d}t + x^3\frac{\mathrm{d}}{\mathrm{d}x}\int_{0}^{x^3+1} e^{t^3} \mathrm{d}t \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \mathrm{d}t + x^3\ \big[3x^2\big] \Bigg[\frac{d}{\mathrm{d}y}\int_{0}^{y} e^{t^3} \mathrm{d}t\Bigg]_{y=x^3+1} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \mathrm{d}t + x^3\ \big[3x^2\big] \Big[e^{y^3}\Big]_{y=x^3+1} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \mathrm{d}t + x^3\ \big[3x^2\big] e^{(x^3+1)^3} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \mathrm{d}t + 3x^5 e^{(x^3+1)^3} \end{aligned}$$

(b) In general, the equation of the tangent line to the graph of $y=f(x)$ at $x=a$ is

$$y=f(a) + f'(a)\,(x-a)$$

Substituting in the given $f(x)$ and $a=-1$:

$$\begin{aligned} f(a)=f(-1)&=(-1)^3\int_0^0 e^{t^3}\ \mathrm{d}t=0\\ f'(a)=f'(-1)&=3(-1)^2\int_{0}^{0}e^{t^3}\ \mathrm{d}t + 3(-1)^5e^{0}\\ &=0-3=-3\\ (x-a)=x-(-1)&=x+1 \\ \text{So, the equation of the tangent line is} \\ y = -3(x+1)\ . \end{aligned}$$

(a) The plot below shows that the rate of pollution appears to be increasing at an increasing rate as time goes on. (b) If we think about an interval on which the rate of pollution is relatively constant and suppose the interval is $\triangle t$ wide, then the area under the curve on that interval is approximately the area of the rectangle whose width is $\triangle t$ and whose height is $p(t)$, the polution rate on that interval. When we compute the product $p(t) \cdot \triangle t$ to find the area, the units involved are “units of pollution per day” times “days”, which results in an overall unit of “units of pollution”. Thus, the area under the curve tells us the total pollution emitted over the time interval $[0,30]$. (c) One way to estimate the total amount of pollution is to suppose the rate of pollution is constant on each of the intervals $[0,6]$, $[6,12]$, $\ldots$. For instance, we could assume the rate on $[0,6]$ is always $7$ units per day, and then the rate jumps to $8$ units per day on $[6,12]$, and so on. Doing so, and using the ideas we discussed in (b), one estimate for the total pollution, $P$, is $$P = 7 \cdot 6 + 8 \cdot 6 + 10 \cdot 6 + 13 \cdot 6 + 18 \cdot 6 = 336$$ total units of pollution. Note, for example, that when we compute $13 \cdot 6$ this represents an estimate of the total amount of pollution emitted on the interval $[18,24]$ since we are assuming the plant emits $13$ units of pollution per day for 6 days. It is apparent that this particular estimate is an underestimate of the total amount of pollution since the rate is always increasing but we are choosing the lowest value of that rate on a given interval.

(a) Since $f(x)$ is positive on $[-1,1]$, the area bounded by the function and $x$-axis on the given interval is $A_1 = \int_{-1}^{1} (3-x^2) \, dx$. A sketch of the function and region is shown below. (b) Likewise, because $g(x)$ is positive on $[-1,1]$, the area bounded by the function and $x$-axis on the given interval is $A_2 = \int_{-1}^{1} 2x^2 \, dx$. A sketch of the function and region is shown below. (c) From the above two figures, we see that the exact area between the two curves is the area under $f$ ($A_1$) minus the area under $g$ ($A_2$). That is, $$A_1 - A_2 = \int_{-1}^{1} (3-x^2) \, dx - \int_{-1}^{1} 2x^2 \, dx\,.$$ (d) By the sum rule for definite integrals over the same interval, we know that $$\int_{-1}^{1} (3-x^2) \, dx - \int_{-1}^{1} 2x^2 \, dx = \int_{-1}^{1} [ (3-x^2) - 2x^2 ] \, dx$$ and thus the latter single integral captures the exact area between the two curves on $[-1,1]$. (e) In general, if $p(x) \ge q(x)$ on $[a,b]$, the exact area, $A$, between $p$ and $q$ on the interval will be given by $$A = \int_a^b [p(x) - q(x)] \, dx$$ because we can find the area as the difference of the area under $p$ ($\int_a^b p(x) \, dx$) and the area under $q$ ($\int_a^b q(x) \, dx$), and then subtract the two definite integrals as shown in (d).

(a) Since the given climb rate, $c(h)$, is in feet per minute, and the new function, $m(h)$, is measured in minutes per foot, it follows that $m(h)$ is the reciprocal of $c(h)$, so $m(h) = \frac{1}{c(h)}$. Hence we get the following updated table.