- Calculus One , Part II
- Week 11 Antidifferentiation
- Week12 Integration
- Week 13 Fundamental Theorem of Calculus
- Week 14 : Substitution Rule
- Week 15: Techniques of Integration
- Week 16: Applications of Integration
You can really think of anti-differentiation as a sort of bridge between the, the differentiation section of this course, and the integration section of this course.
- f(x) = 2x
- F(x) = x²
- G(x) = x²+ 17
- H(x) = x²+ C
- F is an antiderivative of f
∫f(x)dx = F(x)+C
- if ∫f(x)dx = F(x)+C , ∫g(x)dx = G(x)+C
- then ∫(f(x)+g(x))dx = F(x) + G(x) + C ,
- or ∫(f(x)+g(x))dx = ∫f(x)dx + ∫g(x)dx.
- The antiderivative of the sum is the sum of the antiderivative .
- ∫xⁿdx = xⁿ⁺¹/(n+1) + C
f(x) = 15x² -4x +3
∫(15x² -4x +3)dx
= ∫15x²dx - ∫4xdx + ∫3dx
= 15∫x²dx - 4∫xdx + ∫3dx
= 15·x³/3 - 4·x²/2 + 3x + C
= 5x³ - 2x² + 3x + C
- Constant multiple rule:
- ∫a·f(x)dx = a·∫f(x)dx
- The most genral antiderivative of 1/x has the form
⎧ logx + C , if x>0
F(x)= ⎨
⎩ log(-x) + D , if x<0>
- for constant C and D.
- Suppose f is a function with an antiderivative F ,
- Then any another antiderivative for f has the form
- F(x) + C(x)
- for some "locally constant" function C
- C不仅仅是一个常数，C是一个局部常值函数
- 关键在于，这个C可以在不同的区间值不通
- 很不幸， Some textbooks write
- ∫1/xdx = log|x| + C
- is fine provided C is a locally constant function of x.
-
∫cosxdx = sinx + C
-
∫sinxdx = -cosx + C
-
∫tanxdx = log(secx) + C
-
∫secxdx = log|secx + tanx| + C
-
anti-differentiation is HARD.
- ∫eˣdx = eˣ + C
- ∫log(x)dx = x·log(x) - x + C
- if we have ∫f(x)dx = F(x) + C
- ∫f(mx+b)dx = F(mx+b)/m + C
- eg.
- ∫sin(2x+1)dx = -cos(2x+1)/2 + C
- ∫sec²xdx = sin(x)/cos(x)+C , ∫sec²(-5x+7)dx = sin(-5x+7)/cos(-5x+7)/-5 + C = sin(5x-7)/cos(5x-7)/5 +C
- ∫e-x²dx can not be expressed using elementary functions.
- elementary function means polynomials, trig functions, eˣ, log, etc...
- many functions are impossible to antidifferentiate.
- p(t) = position
- v(t) = velocity
- p'(t) = v(t)
- p(t) = ∫v(t)dt
- eg.
- v(t) = 3-10t
- p(t) = ∫(3-10t)dt = 3x - 5t² +C
- and now , the +C has a prefectly reasonable physical interpretation. If I know my velocity, I know my position as long as I know my initial position.
- p(0) = 4
- p(t) = 3x - 5t² + 4
- a(t) = 8
- v(t) = ∫a(t)dt = ∫8dt = 8t + C
- knowning my acceleration doesn't detemine my velociy, it only detemines my velocity up to some constant.
- it could be going really fast or really slow , but still accelerating at the same rate.
- C is v(0)
- p(t) = ∫v(t)dt = ∫(8t + C)dt = ∫8tdt + ∫Cdt = 4t² + Ct + D.
- where C is v(0) , D is p(0)
∫sin²xdx = ∫(1-cos(2x))/2dx
= 1/2∫(1-cos(2x))dx
= 1/2( ∫1dx - ∫cos(2x)dx )
= 1/2( x - sin(2x)/2 ) + C
= x/2 - sin(2x)/4 + C
There's a visual way to gain some insight into these anti-differentiation problems.
- slop field of function x² - x
- instead of plotting a value at some height, I draw little tiny line segments with that slope.
- slop field of function xcosx
- 利用slop field， 可以大致画出 原函数的图像
- graph for y = xsinx + cosx
- +C can move graph up and down.
∑n=ᵏ₁ n = ?
∑n=ᵏ₁ n = (k+1)·k/2
∑n=ᵏ₁ (2n-1)
= ∑n=ᵏ₁ 2n - ∑n=ᵏ₁ 1
= 2·∑n=ᵏ₁ n - ∑n=ᵏ₁ 1
= (k+1)·k - k
= k²
∑n=ᵏ₁ n² = 1² + 2² + ... + k²
= k·(k+1)·(2k+1) /6
大图正好是 4个小图的 3倍。 4个小图平铺，等于大图中间部分。
- the length of buttom of big picture is
2k+1 - the height of big picture is ∑n=ᵏ₁ n =
(k+1)·k/2 - so the sum of small pictures is
(2k+1)·(k+1)·k/6
∑n=ᵏ₁ n³
= ( ∑n=ᵏ₁ n )²
= (k·(k+1)/2)²
= k²·(k+1)² /4
- Thm: If f is continuous, then f is integrable
- means ∫a,bf(x)dx exist.
-
What is the ∫0,1x²dx ?
-
We divide [0,1] into n pieces, so each interval is 1/n
∫0,1x²dx
= limn→∞ ∑i=ⁿ₁ (i/n)²·(1/n)
= limn→∞ ∑i=ⁿ₁ 1/n³·i²
= limn→∞ 1/n³·∑i=ⁿ₁ i²
= limn→∞ ( 1/n³· (n)(n+1)(2n+1)/6 )
= limn→∞ (1/n³·(2n³+3n²+n)/6)
求这个极限很简单
= 1/3
∫0,2x³dx
= limn→∞ ∑i=ⁿ₁ (2/n·i)³·2/n
= limn→∞ ∑i=ⁿ₁ 16/n⁴·i³
= limn→∞ 16/n⁴· ∑i=ⁿ i³
= limn→∞ 16/n⁴· (∑i=ⁿ i)²
= limn→∞ 16/n⁴·((n)(n+1)/2)²
= limn→∞ 4·n²·(n+1)²/n⁴
= 4
repeat this same kind of calculation to deduce that:
∫0,1x³dx = 1/4
now we get the final answer:
∫1,2x³dx = ∫0,2x³dx - ∫0,1x³dx = 15/4
∑n=ᵐ₁ f(n) + ∑n=ᵏm+1 f(n) = ∑n=ᵏ₁ f(n)
∑n=ᵇₐ k·f(n) = k·∑n=ᵇₐf(n)
∑n=ᵇₐ (f(n)+g(n)) = ∑n=ᵇₐ f(n) + ∑n=ᵇₐ g(n)
The derivatives have the same rules about sum !
When is A(x) = ∫ˣₐf(t)dt increasing ? Decreasing ?
The integrals are not exactly measuring area, they're measuring singed area.
- f positive
- A(x) increasing
- A'(x) > 0
- f negative
- A(x) decreasing
- A'(x) < 0
- 利用函数的对称性，我们可以简化 积分计算
- ∫¹₋₁ sinxdx = 0
- ∫2π₀ cosxdx = 0
- Suppose f:[a,b] → ℝ is continuous. let F be the accumulation function , given by
- F(x) = ∫ˣₐ f(t)dt.
- Then F is continuous on [a,b] , differntiable on (a,b) , and F'(x) = f(x)
- So ∫ˣₐ f(t)dt is an antiderivative !
Suppose that ∫³₀ f(x)dx = 9 , and f(3) = 10 , Approximate ∫3.3₀ f(x)dx
∫3.3₀ f(x)dx
= ∫3₀ f(x)dx + ∫3.3₃ f(x)dx
= 9 + 10*0.3 = 12
- we don't case ∫ˣₐ , we really want to calculate ∫ᵇₐ
- F(b) = ∫ᵇₐ f(t)dt
- F(a) = ∫ªₐ f(t)dt = 0
- Suppose f:[a,b] → ℝ is continuous , and F is an antiderivative of f.
- Then ∫ᵇₐ f(x)dx = F(b) - F(a)
- ∫π₀ sinxdx = ?
- -cos(π) - (-cos(0)) = 1 - (-1) = 2
- ∫¹₀ x⁴dx = (1)⁵/5 - (0)⁵/5 = 0.2
- ∫¹₀ (√x -x²)dx = [ x3/2/(3/2) - x³/3 ]¹₀ = 1/3
- F(h) = F(0) + h·F'(0) = F(0) + h·f(0)
- F(2h) = F(h) + h·F'(h) = F(h) + h·f(h) = h·f(0) + h·f(h)
- F(3h) = F(2h) + h·F'(2h) = F(2h) + h·f(2h) = h·f(0) + h·f(h) + h·f(2h)
- ...
- F(10) = h·f(0) + h·f(h) + h·f(2h) + ... + h·f(9h)
- It's Riemann sum!
| Integrating | differentiating |
|---|---|
| summing | differencing |
- 1,2,3,4,5,... -- sum --> 0,1,3,6,10,15,...
- 0,1,3,6,10,15,... -- calc difference --> 1,2,3,4,5,...
- So differences between sum of first k numbers and sum of first k-1 numbers, gives back the original list !
- d/dx ∫ˣₐ f(t)dt = f(x)
- v(t) = my velocity at time t
- ∫ᵇ₀ v(t)dt = distance I traveled t=0 to t=b
- Summarizing , the accumulation function of velocity , is displacement
- The derivative of displacement is velocity.
- we know d/db ∫ᵇₐ f(t)dt = f(b)
- what happens d/da ∫ᵇₐ f(t)dt ?
- -f(a)
- The conventon:
- ∫ᵇₐ f(x)dx = - ∫ªb f(x)dx
- ∫6.02₆ f(x)dx = 0.1 , Approximate f(6) as well as you can given this information
- ∫6.02₆ f(x)dx = f(6) * 0.02 = 0.1
- f(6) = 5
When we first learned about definite integrals, we learned about them as limits of Riemann sum. And in a few cases that definition was good enough. But usually that was much too hard.
So we learned about the fundamental theorem of calculus that reduced evaluating definite integrals down to find anti derivatives.
Now it turns out that finding anti-derivatives is also really hard to do. So we need some better techniques or just heuristics for how to find those anti-derivatives. And a big one is called U-substitution , or maybe the substitution rule. It just running the Chain Rule in reverse.
∫xsin(x²)dx
u = x² , du = 2xdx
I know you might feel kind of bad, because I don't really see a 2xdx, I only see xdx. But his sort of methods going to guide use to do the right thing.
∫xsin(x²)dx = 1/2·∫2xsin(x²)dx = 1/2·∫sin(u)du
= -1/2·cos(u) + C = -1/2·cos(x²) + C
-
Every differentiation rule has a corresponding anti-differentiation rule.
∫f'(g(x))·g'(x)dx , let u=g(x) du=g'(x)dx
= ∫f'(u)du = f(u)+C = f(g(x))+C
-
How to pick u ?
- look for things you can grab as du
- that is , try to find pieces of the integrand that look like the derivative of something.
-
∫x/(√(4-9x²))dx
- u = 4-9x²
-
∫1/(√(4-9x²))dx
- u = 3/2·x
u-substitution is a way to find anti-derivatives. But anti-differentiation is just a means to an end. The real goal, at this point in the course, is evaluating definite integrals.
∫x=²₀ 2x(x²+1)dx
u = x²+1 , du = 2xdx
= ∫x=²₀ u³du = u⁴/4 ］x=²₀
= (x²+1)⁴/4 ］x=²₀
= (2²+1)⁴/4 - (0²+1)⁴/4 = 624/4 = 156
We did it. But I could've finished this problem off in a slightly diferent but equvalent way.
∫x=²₀ u³du
= ∫u=⁵₁ u³du
= u⁴/4 ］u=⁵₁
= 156.
- Method 1 : answer with x
- ∫x=ᵇₐ f'(g(x))g'(x)dx = f(g(x)) ］x=ᵇₐ
- Method 2 : endpoints with u
- ∫x=ᵇₐ f'(g(x))g'(x)dx = ∫u=g(b)g(a) f'(u)du = f(u) ］u=g(b)g(a)
-
Sometimes you might want to do u substitution more than once.
∫-2cosx sinx cos(cos²x+1)dx
u= cosx , du = -sinxdx
= ∫2u cos(u²+1) du
v = u²+1 , dv = 2udu
= ∫cosvdv = sinv + C
= sin( u²+1 ) + C
= sin( cos²x +1 ) + C
We know ∫1/(1+x²)dx = atan(x) + C
The trick is completing the square.
x² + 4x + 7
= (x+2)² + 3
So :
∫1/(x² + 4x + 7)dx
= ∫1/((x+2)² + 3 )dx
= 1/3·∫1/(1/3·(x+2)² + 1)dx
let u= 1/√3·(x+2) , du = 1/√3dx
= 1/√3·∫1/(1/3·(x+2)² + 1)·(1/√3)dx
= 1/√3·∫ 1/(u²+1)du
= 1/√3·atan(u) + C
= ...
∫x=¹₀ (x+10)(x-1)¹⁰dx
let u =x-1, du = dx , x = u+1
= ∫x=¹₀ (u+11)u¹⁰du
= ∫x=¹₀ u¹¹+11u¹⁰du
= u¹²/12 + u¹¹ ］u=⁰₋₁
= 11/12
∫x/∛(x+1)dx
let u=x+1, du=dx , x=u-1
= ∫(u-1)/∛u du
There is a more easy way to do this.
let u = ∛(x+1) , u³=x+1, x=u³-1 , dx = 3u²du
= ∫(u³-1)/u·3u²du
= ∫(3u⁴-3u)du // it is polynomial
= 3/5·u⁵ - 3/2·u² + C
- This is called rationalizing substitution.
Sometimes the best substitution to make isn't even visible until after we've messed around with the integrand some how.
∫ 1/(1+cosx) dx
= ∫ 1/(1+cosx)·(1-cosx)/(1-cosx) dx
= ∫ (1-cosx)/sin²x dx
= ∫ 1/sin²xdx - ∫ cosx/sin²xdx
= -cotx - ∫ cosx/sin²xdx
let u = sinx, du=cosxdx
= -cotx - ∫ 1/u²du
= -cotx - 1/u +C
= -cotx - 1/sinx +C
What is d/dx ∫x²₀ sintdt ? How to deal with x² ?
We know d/dx ∫x₀ sintdt = sinx. What I'm asking is what if this endpoint weren't x anymore , but some funtion g(x) ? ∫g(x)₀ sintdt = ?
f(x) = ∫x₀ sintdt
=> f(g(x)) = ∫g(x)₀ sintdt
=> d/dx f(g(x)) = d/dx ∫g(x)₀ sintdt
d/dx f(g(x))
= f'(g(x))·g'(x)
f(x) = ∫x₀ sintdt
=> f'(x) = sinx
=> d/dx f(g(x)) = sin(g(x))·g'(x)
g(x) = x² , g'(x) = 2x
=> d/dx f(g(x)) = sin(x²)·2x
Define a functin f: ℝ → ℝ , by the rule f(t) = ∫sint₀ cosxdx. what is f'(t) ?
- (cost)(cossint)
F(x) = ∫ˣₐf(t)dt
F'(x) = f(x)
- a, x, t
∫d/dx(f(x)g(x))dx = ∫(f'(x)g(x) + f(x)g'(x))dx = f(x)g(x) + C
∫f'(x)g(x)dx + ∫f(x)g'(x)dx = f(x)g(x) + C
∫f(x)g'(x)dx = f(x)g(x) - ∫f'(x)g(x)dx
- what this is saying ?
- It's saying that I can do
∫f(x)g'(x)dxif I can do∫f'(x)g(x)dx
- It's saying that I can do
let u=f(x), dv = g'(x)dx
du = f'(x), v=g(x)
so we get :
∫udv = uv - ∫vdu.
This is maybe why it makes sense to call this integration by parts.
It's trading game. I'm trading ∫udv with ∫vdu. But now on part is differentiated and another part of the inner grand is antidifferentiated.
The basic idea of integration by parts is that it lets you differentiate part of the integrand, but only if you're willing to pay a price. And that price is anti-differentiating the other part of the integrand.
∫xeˣdx
let u=x, dv=eˣdx
so du = dx , v = eˣ , now we get
∫xeˣdx = xeˣ - ∫eˣdx = xeˣ - eˣ + C
Now, we can use the same trick to attack similar integration problems. For example, let's say you want to anti-differentiate some polynomial in x times eˣ.
∫logxdx
let u=logx, dv = dx , so du=1/xdx , v = x
so
∫logxdx = xlogx - ∫x·1/xdx = xlogx - x + C
let u=eˣ , dv = cosxdx
so ∫eˣcosxdx = eˣsinx - ∫sinxeˣdx
what is ∫sinxeˣdx ?
let u=eˣ , dv = sinxdx
we get ∫sinxeˣdx = -eˣcosx + ∫cosxeˣdx
=>
∫eˣcosxdx = eˣsinx - ( -eˣcosx + ∫cosxeˣdx )
=> 2·∫eˣcosxdx = eˣsinx + eˣcosx
=> ∫eˣcosxdx = (eˣsinx + eˣcosx )/2
Sometimes it isn't clear that you can do integration by parts until after you perform some substitution
∫e√xdx
let u=√x , du = 1/(2√x)dx , dx = 2√xdu
∫e√xdx = ∫eᵘ·2udu
let v=2u , dw = eᵘdu ; dv = 2du , w = eᵘ
注意这里的选择很重要 eᵘdu 可以极大简化运算
= 2u·eᵘ - ∫eᵘ2du
= 2u·eᵘ - 2eᵘ
= 2√x·e√x - 2e√x
= 2e√x (√x-1)
-
trick: You can trade sines for cosines or vice versa.
- sin²x + cos²x = 1
∫sin³x·cos²xdx
= ∫sinx·sin²x·cos²xdx
= ∫sinx·(1-cos²x)·cos²xdx
let u=cosx, du=-sinxdx
= -∫(1-u²)·u²du
= -( u³/3 - u⁵/5 ) + C
= -cos³x/3 + cos⁵x/5 + C
- The trick works as long as we've got an odd power on the sine, or an odd power on the cosine.
- the key is that you can convert that odd power to
even powerx (sin/cos) - and the
even powercan apply the rulesin²x + cos²x = 1
- the key is that you can convert that odd power to
Now we know how the handle odd power of sin/cos. But how to handle even power ? We need the Half·Angle Formual.
- Half·Angle Formual
- sin²x = (1-cos(2x))/2
- cos²x = (1+cos(2x))/2
∫sin⁴xdx
= ∫(sin²x)²dx
= ∫( (1-cos(2x))/2 )²dx
= ∫( 1/4 + cos²(2x)/4 - cos(2x)/2 )dx
= ∫cos²(2x)/4dx - ∫cos(2x)/2dx + ∫1/4dx
= ∫cos²(2x)/4dx - sin(2x)/4 + 1/4x
apply Half·Angle Formual again
= ∫(1+cos(4x))/8dx - sin(2x)/4 + 1/4x
= 1/8x + sin(4x)/32 - sin(2x)/4 + 1/4x +C
= 3/8x + sin(4x)/32 - sin(2x)/4 + C
I should say that in some cases you can get away with doing a bit less work.
∫π₀ sin²ⁿxdx
// cos(2x) would integrate to 0. but cos²(2x) doesn't !
= ∫π₀ ( 1/4 + cos²(2x)/4 - cos(2x)/2 )dx
= ∫π₀ ( 1/4 + (1+ cos(4x) )/8 ) dx
= ∫π₀ 3/8 dx = 3/8 π
∫π/2₀ sin³²xdx = 300540195·π / 4294967296
= 3*3*5*17*19*23*29*31* π / 2³²
Why does it factor so nicely ?
TODO
- height of rectangle : dy
- width of rectangle: 2√y
- area = ∫¹₀ 2√ydy (Ps. not [-1,1], [-1,1] is for ∫xxxdx )
the bad news about vertical strips is that there's 2 different kinds of vertical strips.
The vertical strips on the left bottom-left side touch the orange and the purple edge, and the vertical strips on the right-top touch orange and blue edge.
if I cut this up into horizontal strips, all of my horizontal strips have an orange side and a blue side. I have only one type of horizontal strip.
- height: dy
- width : (y²+1)/2 - y²
- area = ∫¹₀ ((y²+1)/2 - y²) dy
Write an integral representing the area of the region bounded by the lines x=8, x=12, the function f(x)=6x+7, and the function g(x) = √(10+arctan(x)).
∫¹²₈ |√(10+arctan(x) - (6x+7))|dx (注意绝对值号, 这里是求面积，而不是积分的有向面积)
cut this thing up into slices and add up the volumes of the little slices
And what is the volumn of one of those slice ?
= πr²dx
volumn = ∫ₓ₌²₀ π(x/2)²dx
= π/4·∫ₓ₌²₀ x²dx = π/4·x³/3 ]²₀
= 2π/3
volumn = ∫ʳ₋ᵣ π(√(r²-x²))²dy
= π∫ʳ₋ᵣ (r²-x²)dy
= 2π ( r²y - y³/3 ］ʳ₀ )
= 2π ( r²r - r³/3 )
= 4/3πr³
计算绿色方块所在图像，饶 x 轴旋转一周后，所形成的物体的体积。
We pick one of the green rectangle, and rotate 360' by x. We call this object "washer".
What's the volumn of a washer ?
v_washer = π(R²-r²)dx
and we know:
R = √x , r = x²
so
V = ∫¹₀ π(√x²-(x²)²)dx
= π∫¹₀ (x-x⁴)dx
= π( x²/2 - x⁵/5 )¹₀
= 0.3π
for each shell :
- h = 1-x²
- 横截面面积 s = πR²-πr² = π(R+r)(R-r) = π·2x·dx
V = ∫¹₀ π·2x·(1-x²) dx
= π/2
we can solve it by cutting it in shells or washers.
we use shells here.
V = ∫²₁ 2πx·2√(4-x²)dx
= 4π·√3
The total length of stairs is 2 . What if I make those stairs even smaller ? still 2.
What if I make those stairs infinitely smaller ?
Let us ask another question: how to know the length of a curve y = f(x) ?
for every piece of that curve , the length is √(dx²+dy²) , after I integrate , the length of the curve is :
∫ (√(dx²+dy²))
now we pulling the dx outside
= ∫ √(1+(dy/dx)²)dx
So here's the formula that we're going to use ?
- the arc length from (a,f(a)) to (b,f(b)) along the graph of y=f(x) is :
- ∫ᵇₐ √(1+f'(x)²) dx
dy/dx = 3/2·x1/2
so length = ∫¹₀ √(1+(3/2·x1/2)²) dx
= ∫¹₀ √(1+9/4·x) dx
= (13·√13)/27 - 8/27