Lesson 11-18 True/False
Any polynomial can be factored as a product of linear factors (some of which may possibly be repeated), which proves that the method of partial fractions will always work for integrals like ∫𝑃(𝑥)𝑄(𝑥)𝑑𝑥∫P(x)Q(x)dx.
F
Checking your answers after integrating by substitution in this HW is a waste of time.
F
FTC Part 2 says that if you start with a function 𝐹(𝑥)F(x), then differentiate it, and then integrate it (assuming all these operations are OK), you will get back the original function 𝐹(𝑥)F(x).
F
FTC Part 2 says that ∫2−11𝑥𝑑𝑥=ln2−ln1.
F
Finding the intersection points of the graphs of two functions 𝑦=𝑓(𝑥)y=f(x) and 𝑦=𝑔(𝑥)y=g(x) is crucial when we are trying to find the area bounded by the curves, but it is not necessary when an interval [𝑎,𝑏][a,b] is specified.
F
If a solid of revolution is made of the material between two functions 𝑦=𝑓(𝑥)y=f(x) and 𝑦=𝑔(𝑥)y=g(x) on [𝑎,𝑏][a,b] that are both revolved about the 𝑥x-axis, then its volume is given by 𝜋∫𝑏𝑎(𝑓(𝑥)−𝑔(𝑥))2𝑑𝑥π∫ab(f(x)−g(x))2dx, while the area between the two functions is given by ∫𝑏𝑎|𝑓(𝑥)−𝑔(𝑥)|𝑑𝑥.
F
If a solid of revolution is obtained by rotating a function 𝑦=𝑓(𝑥)y=f(x) about the 𝑥x-axis, the formula for its volume can be viewed as the sum of the volumes of infinitely many cylinders of height 𝑑𝑥dx.
F
If the denominator 𝑄(𝑥)Q(x) in ∫𝑃(𝑥)𝑄(𝑥)𝑑𝑥∫P(x)Q(x)dx is of degree 3, correctly setting up a partial fractions solution leads to a system of 4 equations and 4 unknowns, essentially because a polynomial of degree 3 has 4 coefficients to equate.
F
If the numerator 𝑃(𝑥)P(x) in ∫𝑃(𝑥)𝑄(𝑥)𝑑𝑥∫P(x)Q(x)dx can be factored into distinct linear factors (𝑥−𝑎𝑖)(x−ai), then we will not need the method of equating coefficients to determine partial fractions, because we will have enough roots 𝑎𝑖ai to plug in.
F
If you are trying to approximate the area under a curve 𝑦=𝑓(𝑥)y=f(x), then a Riemann sum with 1001 rectangles will definitely give you a better answer than a Riemann sum with 10 rectangles.
F
If you take a Riemann sum with MANY rectangles, say a billion, then that will give you the true net area you are looking for.
F
If ∫∞𝑎𝑓(𝑥)𝑑𝑥∫a∞f(x)dx and ∫∞𝑎𝑔(𝑥)𝑑𝑥∫a∞g(x)dx are both divergent, then ∫∞𝑎[𝑓(𝑥)+𝑔(𝑥)]𝑑𝑥∫a∞[f(x)+g(x)]dx is also divergent
F
If ∫∞𝑎𝑓(𝑥)𝑑𝑥∫a∞f(x)dx and ∫∞𝑎𝑔(𝑥)𝑑𝑥∫a∞g(x)dx are both divergent, then ∫∞𝑎[𝑓(𝑥)𝑔(𝑥)]𝑑𝑥∫a∞[f(x)g(x)]dx is also divergent.
F
If 𝑓 is a continuous decreasing function on [1,∞)[1,∞) and lim𝑡→∞𝑓(𝑥)=0limt→∞f(x)=0, then ∫∞1𝑓(𝑥)𝑑𝑥∫1∞f(x)dx is convergent.
F
If 𝑓f and 𝑔g are continuous on [𝑎,𝑏][a,b], then ∫𝑏𝑎[𝑓(𝑥)𝑔(𝑥)]𝑑𝑥=(∫𝑏𝑎𝑓(𝑥)𝑑𝑥)(∫𝑏𝑎𝑔(𝑥)𝑑𝑥)
F
If 𝑓f and 𝑔g are differentiable and 𝑓(𝑥)≥𝑔(𝑥)f(x)≥g(x) for all 𝑎<𝑥<𝑏a<x<b, then 𝑓′(𝑥)≥𝑔′(𝑥)f′(x)≥g′(x) for 𝑎<𝑥<𝑏a<x<b.
F
If 𝑓f is continuous on [1,3][1,3], then we must have ∫31𝑓′(𝑣)𝑑𝑣=𝑓(3)−𝑓(1)∫13f′(v)dv=f(3)−f(1).
F
If 𝑓f is continuous on [𝑎,𝑏][a,b], then 𝑑𝑑𝑥(∫𝑏𝑎𝑓(𝑥)𝑑𝑥)=𝑓(𝑥)ddx(∫abf(x)dx)=f(x).
F
If 𝑓f is continuous on the interval [𝑎,𝑏][a,b], then ∫𝑏𝑎𝑥𝑓(𝑥)𝑑𝑥=𝑥∫𝑏𝑎𝑓(𝑥)𝑑𝑥∫abxf(x)dx=x∫abf(x)dx.
F
If 𝑓f is continuous, then ∫∞−∞𝑓(𝑥)𝑑𝑥=lim𝑡→∞∫𝑡−𝑡𝑓(𝑥)𝑑𝑥∫−∞∞f(x)dx=limt→∞∫−ttf(x)dx.
F
If 𝑔(𝑥)≤𝑓(𝑥)g(x)≤f(x) on [0,∞)[0,∞) and ∫∞0𝑓(𝑥)𝑑𝑥∫0∞f(x)dx converges, then ∫∞0𝑔(𝑥)𝑑𝑥∫0∞g(x)dx also converges.
F
Integration by parts is the best and only way to compute the integral ∫sin(𝑥)cos(𝑥)𝑑𝑥∫sin(x)cos(x)dx.
F
It is impossible to compute the exact value of ∫∞−∞𝑒−𝑥2𝑑𝑥∫−∞∞e−x2dx because there is no antiderivative for 𝑓(𝑥)=𝑒−𝑥2f(x)=e−x2 in terms of elementary functions.
F
Long division of polynomials can be avoided even in the case when the numerator has a degree larger than the degree of the denominator, as long as we can somehow find the unknown constants in the numerators of the corresponding partial fractions.
F
Since applying integration by parts to ∫𝑒𝑥sin(𝑥)𝑑𝑥∫exsin(x)dx with 𝑢=sin(𝑥),𝑣′=𝑒𝑥u=sin(x),v′=ex produces 𝑒𝑥sin(𝑥)−∫𝑒𝑥cos(𝑥)𝑑𝑥exsin(x)−∫excos(x)dx, integration by parts isn't helpful for this integral.
F
Splitting an integral along its interval as in ∫𝑏𝑎𝑓(𝑥)𝑑𝑥=∫𝑐𝑎𝑓(𝑥)𝑑𝑥+∫𝑏𝑐𝑓(𝑥)𝑑𝑥∫abf(x)dx=∫acf(x)dx+∫cbf(x)dx makes sense only when 𝑐c is between 𝑎a and 𝑏b.
F
The 'net-area-so-far' function 𝐹(𝑥)=∫𝑥𝑎𝑓(𝑢)𝑑𝑢F(x)=∫axf(u)du is positive when 𝑓(𝑥)f(x) increases, and negative when 𝑓(𝑥)f(x) decreases.
F
The definite integral ∫20(𝑥−𝑥3)𝑑𝑥∫02(x−x3)dx represents the area under the curve 𝑦=𝑥−𝑥3y=x−x3 and above the 𝑥x-axis, on the interval 0≤𝑥≤20≤x≤2.
F
The formula for the volume of a solid whose cross-sections perpendicular to the 𝑥x-axis have area 𝐴(𝑥)A(x) is a special case of the formula for the volume of a solid of revolution.
F
The formula 𝐴=∫𝑏𝑎|𝑓(𝑥)−𝑔(𝑥)|𝑑𝑥A=∫ab|f(x)−g(x)|dx for the total area between the curves 𝑦=𝑓(𝑥)y=f(x) and 𝑦=𝑔(𝑥)y=g(x) on [𝑎,𝑏][a,b] has to be modified in case one or both curves are below the 𝑥x-axis.
F
The fraction 𝑥(𝑥2+3)𝑥3−8x(x2+3)x3−8 can be written as 𝐴𝑥−2+𝐵𝑥+𝐶𝑥2+2𝑥+4Ax−2+Bx+Cx2+2x+4 for some constants 𝐴A and 𝐵B.
F
The method of partial fractions cannot be used to compute the integral ∫(cos𝑥)/sin3𝑥+sin𝑥𝑑𝑥∫cosxsin3x+sinxdx, since there are no polynomials in the numerator or denominator of the integrand.
F
To calculate the definite integral ∫5−525−𝑥2‾‾‾‾‾‾‾√𝑑𝑥∫−5525−x2dx, we must find an antiderivative of 25−𝑥2‾‾‾‾‾‾‾√25−x2 and evaluate it at the ends of the interval [−5,5][−5,5].
F
To find an antiderivative of a quotient, you can simply divide any antiderivative of the numerator by any antiderivative of the denominator.
F
We can rewrite the definite integral ∫50𝑥1+𝑥2‾‾‾‾‾‾√𝑑𝑥∫05x1+x2dx using substitution with 𝑢=1+𝑥2u=1+x2and 𝑑𝑢=2𝑥𝑑𝑥du=2xdx as ∫50𝑢√2𝑑𝑢∫05u2du.
F
We can show that ∫∞51𝑥1.01𝑑𝑥∫5∞1x1.01dx converges in at least three ways: by a brute force calculation using the definition of an improper integral; by representing ∫∞51𝑥1.01𝑑𝑥∫5∞1x1.01dx as part of ∫∞11𝑥1.01𝑑𝑥∫1∞1x1.01dx and then using a formula from class for the value of the latter integral; or by comparing it with the more familiar integral ∫∞51𝑥1𝑑𝑥∫5∞1x1dx.
F
We cannot compute the integral ∫19−16𝑥2‾‾‾‾‾‾‾‾√𝑑𝑥∫19−16x2dx with the integral techniques we have learned so far.
F
We need to apply integration by parts three times to find ∫[ln(𝑥)]2𝑑𝑥∫[ln(x)]2dx.
F
When asked for the area of a region, it's fine to give a negative answer if it is bounded above by the 𝑥x-axis and below by some function 𝑓(𝑥)f(x).
F
When determining whether ∫∞−∞𝑓(𝑥)𝑑𝑥∫−∞∞f(x)dx converges, we must split the integral at 0 as ∫0−∞𝑓(𝑥)𝑑𝑥+∫∞0𝑓(𝑥)𝑑𝑥∫−∞0f(x)dx+∫0∞f(x)dx and check convergence for the two integrals separately.
F
When solving an integral ∫𝑥ℎ(𝑥)𝑑𝑥∫xh(x)dx with integration by parts, you should always pick 𝑢=𝑥u=x and 𝑣′=ℎ(𝑥)v′=h(x).
F
When we do not know an antiderivative of a function and we cannot find the limit of the corresponding Riemann sums on [𝑎,𝑏], searching for a geometric interpretation of the desired definite integral is also useless.
F
(Challenge) Riemann sums are somewhat cumbersome tools for finding approximations of areas, yet they are absolutely necessary to link antiderivatives to areas.
T
A reasonable first thing to try on a substitution problem is to let 𝑢u be the inside function of a composition, e.g. a variable expression being raised to a constant power, a variable expression as the exponent for a constant base, the denominator of a fraction, or the input to a log function.
T
All continuous functions have antiderivatives.
T
Antiderivatives can be useful in at least three places: solving simple differential equations, finding areas of shapes in the plane, and finding distances traveled by objects.
T
At this point, we have not covered a technique for finding an antiderivative of 𝑦=tan(𝑥)y=tan(x).
T
Bounding 𝑓(𝑥)f(x) on [𝑎,𝑏][a,b] from above and below by some constants 𝑀M and 𝑚m will, in general, produce only an estimate of ∫𝑏𝑎𝑓(𝑥)𝑑𝑥∫abf(x)dx, not the exact value.
T
Calculating the volume of a sphere as a solid of revolution is simpler than calculating the area of a circle using integration.
T
Due to symmetry, we can see that ∫1−1(𝑥5−6𝑥9+sin𝑥(1+𝑥4)2)𝑑𝑥=0∫−11(x5−6x9+sinx(1+x4)2)dx=0 without taking any antiderivatives.
T
Even though the name 'initial value problem' probably comes from specifying the output at time 𝑡=0t=0, it's not actually mandatory to specify 𝑦(0)y(0); a different point could be given.
T
FTC Part 1 says that if you start with a function 𝑓(𝑥)f(x), then integrate 𝑓(𝑢)f(u) from 𝑎a to 𝑥x with respect to 𝑢u, and finally differentiate with respect to 𝑥x (assuming all these operations are OK), then you will get back the original function 𝑓(𝑥)f(x).
T
If a function has at least one antiderivative, then it has infinitely many antiderivatives.
T
If the denominator 𝑄(𝑥)Q(x) in ∫𝑃(𝑥)𝑄(𝑥)𝑑𝑥∫P(x)Q(x)dx can be factored as a product of distinct linear factors, it is almost certainly more efficient to choose clever values of the variable than to match coefficient on the polynomial equation. [Adjusted 10/6 to match bCourses.]
T
If the denominator 𝑄(𝑥)Q(x) in ∫𝑃(𝑥)𝑄(𝑥)𝑑𝑥∫P(x)Q(x)dx can be factored as a product of linear factors, but some factors are repeated, we can still plug in values to determine some of the coefficients of the partial sum quickly.
T
If two functions have the same family of antiderivatives, then they must actually be the same function.
T
If we are given 𝑓″(𝑥)=2+𝑥3+𝑥5f″(x)=2+x3+x5 and no other information, then we can find the most general form for 𝑓f, but there will be two arbitrary constants in our answer.
T
If you didn't do as well as you'd like on the midterm, it would be a good idea to go to office hours more.
T
If ∫∞𝑎𝑓(𝑥)𝑑𝑥∫a∞f(x)dx and ∫∞𝑎𝑔(𝑥)𝑑𝑥∫a∞g(x)dx are both convergent, then ∫∞𝑎[𝑓(𝑥)+𝑔(𝑥)]𝑑𝑥∫a∞[f(x)+g(x)]dx is also convergent.
T
If 𝑓f and 𝑔g are continuous and 𝑓(𝑥)≥𝑔(𝑥)f(x)≥g(x) for 𝑎<𝑥<𝑏a<x<b, then ∫𝑏𝑎𝑓(𝑥)𝑑𝑥≥∫𝑏𝑎𝑔(𝑥)𝑑𝑥∫abf(x)dx≥∫abg(x)dx
T
Integrals can be used to derive or come up with some familiar geometry formulas, such as the area of a trapezoid.
T
Integration by parts is essentially an integration law for products because it 'undoes' for antiderivatives what the Product Rule does for derivatives.
T
It is essential to use the correct differential (e.g. 𝑑𝑥dx or 𝑑𝑢du) when doing integration.
T
Just as writing lim in your limit work was required until the final evaluation step, you must write the differential 𝑑𝑥dx in your integration work until the point where you have evaluated or applied FTC Part 2.
T
The '3' in the denominator for the volume of a sphere essentially comes from integrating 𝑥^2.
T
The 'net-area-so-far' function 𝐹(𝑥)=∫𝑥𝑎𝑓(𝑢)𝑑𝑢F(x)=∫axf(u)du is concave up where 𝑓(𝑥)f(x) is increasing, and it is concave down where 𝑓(𝑥)f(x) is decreasing.
T
The Substitution Rule is essentially an integration law for composition because it 'undoes' for antiderivatives what the Chain Rule does for derivatives.
T
The area between the curves 𝑦=𝑥2y=x2 and 𝑦=𝑥2+2y=x2+2 for −1≤𝑥≤3−1≤x≤3 is the same as the area of a rectangle of side lengths 2 and 4.
T
The fact that the method of partial fractions always works on any integral of the type ∫𝑃(𝑥)𝑄(𝑥)𝑑𝑥∫P(x)Q(x)dx fundamentally depends on being able to solve systems of n linear equations with n unknowns, which in turn is a central topic in Linear Algebra, to be briefly studied later in the semester in MATH 10A.
T
The indefinite integral ∫25−𝑥2‾‾‾‾‾‾‾√𝑑𝑥∫25−x2dx can be found using the same IBP method as finding ∫sin𝑥⋅𝑒𝑥𝑑𝑥∫sinx⋅exdx.
T
The number of unknown constants in a properly-set-up partial fractions sum is at most the degree of the denominator in the original fraction. [Adjusted 10/6 to match bCourses.]
T
There are at least two ways to compute ∫𝜋−𝜋sin𝑥𝑑𝑥∫−ππsinxdx.
T
There are two different ways to calculate definite integrals with the substitution rule: (a) temporarily forget about the bounds of integration, find an antiderivative, and then use FTC Part II; or (b) adjust the bounds of integration as you substitute to reflect the change of variables.
T
To find an antiderivative of a sum, you can find the sum of the individual antiderivatives (provided all necessary antiderivatives exist).
T
To show that the claim 'The integral of a product is the product of the two integrals' is flawed, it suffices to produce one counterexample where it does not work.
T
We are able to find an antiderivative of 𝑓(𝑥)=𝑥4−𝑥2+1𝑥2f(x)=x4−x2+1x2 using techniques and laws from this lesson, even though 𝑓f is defined using a quotient.
T
We can use even and odd symmetry to show that ∫5−5(𝑎𝑥2+𝑏𝑥+𝑐)𝑑𝑥=2∫50(𝑎𝑥2+𝑐)𝑑𝑥∫−55(ax2+bx+c)dx=2∫05(ax2+c)dx.
T
We explained the formula 43𝜋𝑟343πr3 for the volume of a sphere of radius 𝑟 by picturing the sphere as a solid of revolution.
T
When computing improper integrals, you must leave the limits in your work until you are able to evaluate them.
T
When solving an integral ∫ℎ1(𝑥)ℎ2(𝑥)𝑑𝑥∫h1(x)h2(x)dx with integration by parts, you should usually pick 𝑣′v′ to be the function with the simpler antiderivative.
T
