1S11: Calculus for students in Science

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Dr. Vladimir Dotsenko

TCD

Michaelmas Term 2013

Dr. Vladimir Dotsenko (TCD) 1S11: Calculus for students in Science Michaelmas Term 2013 1 / 13

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Recall the u-substitution method for computing antiderivatives: given an integral of the form

Z

f(g(x))g^′(x)dx,

we denote u =g(x) so thatdu =g^′(x)dx, so that the integral becomes Z

f(u)du.

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Z

f(g(x))g^′(x)dx,

f(u)du.

In order to use this method to evaluate definite integrals of the same form Z b

a

f(g(x))g^′(x)dx,

we need to take appropriate care of the effect of that on the limits of

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Z

f(g(x))g^′(x)dx,

f(u)du.

In order to use this method to evaluate definite integrals of the same form Z b

a

f(g(x))g^′(x)dx,

we need to take appropriate care of the effect of that on the limits of integration. There are two ways to deal with it, which we shall now outline.

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Method 1. Useu-substitutions only on the level of indefinite integrals:

evaluate Z

f(g(x))g^′(x)dx, and then use the formula

Z b a

f(g(x))g^′(x)dx = Z

f(g(x))g^′(x)dx b

a

.

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Method 1. Useu-substitutions only on the level of indefinite integrals:

evaluate Z

f(g(x))g^′(x)dx, and then use the formula

Z b a

f(g(x))g^′(x)dx = Z

f(g(x))g^′(x)dx b

a

.

Method 2. Use the relationshipu =g(x) to modify the limits:

Z b a

f(g(x))g^′(x)dx = Z g(b)

g(a)

f(u)du.

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Example 1. Let us use the first method to evaluate Z 2

0

x(x²+ 1)³dx.

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0

x(x²+ 1)³dx. We denote u=x²+ 1, so thatdu = 2x dx, and

Z

x(x²+ 1)³dx = 1 2

Z

u³du = u⁴

8 +C = (x²+ 1)⁴ 8 +C.

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0

x(x²+ 1)³dx. We denote u=x²+ 1, so thatdu = 2x dx, and

Z

x(x²+ 1)³dx = 1 2

Z

u³du = u⁴

8 +C = (x²+ 1)⁴ 8 +C. Therefore,

Z 2 0

x(x²+ 1)³dx = Z

x(x²+ 1)³dx 2

0

=

(x²+ 1)⁴ 8

2 0

= 625 8 −1

8 = 78.

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Example 2. Let us use the second method to evaluate the same integral Z ₂

0

x(x²+ 1)³dx.

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0

x(x²+ 1)³dx. We denote u =x²+ 1, so thatdu= 2x dx, and

for x= 0,u = 1, for x= 2,u = 5.

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0

x(x²+ 1)³dx. We denote u =x²+ 1, so thatdu= 2x dx, and

for x= 0,u = 1, for x= 2,u = 5.

Therefore, Z ₂

0

x(x²+ 1)³dx = 1 2

Z ₅

1

u³du = 1 2

5⁴ 4 −1⁴

4

= 78.

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Example 3. Let us evaluate the integral Z ₃

1

cos(π/x) x² dx.

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1

cos(π/x) x² dx.

We putu= ^π_x, so that du=−_x^π²dx, in other words, _x¹2dx =−_π¹du.

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1

cos(π/x) x² dx.

We note that

for x = 1,u =π, for x= 3,u =π/3.

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1

cos(π/x) x² dx.

We note that

for x = 1,u =π, for x= 3,u =π/3.

Therefore, Z 3

1

cos(π/x)

x² dx =−1 π

Z π/3 π

cosu du=

=− 1 π sinu

π/3 π

=−1

π(sin(π/3)−sinπ) =−

√3

2π ≈ −0.276.

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Example 4. Let us evaluate the integral Z _π/4

0

√tanx 1 cos²xdx.

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0

√tanx 1 cos²xdx. We putu= tanx, so thatdu= _cos¹₂_x dx.

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0

√tanx 1 cos²xdx.

We putu= tanx, so thatdu= _cos¹₂_x dx. We note that for x= 0,u = 0,

for x=π/4,u = 1.

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0

√tanx 1 cos²xdx.

We putu= tanx, so thatdu= _cos¹₂_x dx. We note that for x= 0,u = 0,

for x=π/4,u = 1.

Therefore, Z _π/4

0

√tanx 1

cos²x dx = Z ₁

0

√u du= u^3/2 3/2

#1

0

= 2 3.

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Example 5. Let us prove, without evaluating integrals, that Z π/2

0

sinⁿx dx = Z π/2

0

cosⁿx dx.

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0

sinⁿx dx = Z π/2

0

cosⁿx dx.

In the second integral, we putu = ^π₂ −x, so thatdu=−dx.

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0

sinⁿx dx = Z π/2

0

cosⁿx dx.

In the second integral, we putu = ^π₂ −x, so thatdu=−dx. We note that

for x= 0,u =π/2, for x=π/2,u = 0, and that cosx= cos(π/2−u) = sinu.

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0

sinⁿx dx = Z π/2

0

cosⁿx dx.

In the second integral, we putu = ^π₂ −x, so thatdu=−dx. We note that

for x= 0,u =π/2, for x=π/2,u = 0, and that cosx= cos(π/2−u) = sinu. Therefore,

Z _π/2

0

cosⁿx dx =− Z ₀

π/2

sinⁿu du= Z _π/2

0

sinⁿx dx, as required.

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Example 6. Let us examine the integral Z ₁

−1

1 1 +x² dx.

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−1

1 1 +x² dx.

Let us perform the substitutionu= _x¹, so that du=−_x¹2dx, in other words, du=−u²dx anddx =−_u¹2du.

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−1

1 1 +x² dx.

Let us perform the substitutionu= _x¹, so that du=−_x¹2dx, in other words, du=−u²dx anddx =−_u¹2du. We note that

forx =−1,u =−1, and for x = 1,u = 1, and that _1+x¹ 2 = _1+(1/u)¹ 2 = _1+u^u²2.

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−1

1 1 +x² dx.

forx =−1,u =−1, and for x = 1,u = 1, and that _1+x¹ 2 = _1+(1/u)¹ 2 = _1+u^u²2. Therefore,

Z 1

−1

1

1 +x² dx =− Z 1

−1

u² 1 +u²

1

u² du=− Z 1

−1

1 1 +u² du, so the integral is equal to its negative and hence equal to zero.

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−1

1 1 +x² dx.

Z 1

−1

1

1 +x² dx =− Z 1

−1

u² 1 +u²

1

u² du=− Z 1

−1

1 1 +u² du, so the integral is equal to its negative and hence equal to zero. How is it possible?

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−1

1 1 +x² dx.

Z 1

−1

1

1 +x² dx =− Z 1

−1

u² 1 +u²

1

u² du=− Z 1

−1

1 1 +u² du, so the integral is equal to its negative and hence equal to zero. How is it

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Example 7. (High school maths intervarsity competitions in Russia) Let us evaluate the integral

Z π/2 0

(sin²(sinx) + cos²(cosx))dx.

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Z π/2 0

(sin²(sinx) + cos²(cosx))dx. We shall split this integral as a sum

Z π/2 0

sin²(sinx)dx+ Z π/2

0

cos²(cosx)dx,

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Z π/2 0

0

cos²(cosx)dx,

and transform the second integral using the substitutionu = ^π₂ −x, so that du=−dx,

forx = 0,u =π/2, for x =π/2,u = 0, cos(x) = cos(π/2−u) = sinu

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Z π/2 0

0

cos²(cosx)dx,

and transform the second integral using the substitutionu = ^π₂ −x, so that du=−dx,

forx = 0,u =π/2, for x =π/2,u = 0, cos(x) = cos(π/2−u) = sinu, leading to

Z _π/2

0

cos²(cosx)dx =− Z ₀

π/2

cos²(sinu)du= Z _π/2

0

cos²(sinx)dx.

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Therefore, Z _π/2

0

(sin²(sinx) + cos²(cosx))dx =

= Z _π/2

0

sin²(sinx)dx+ Z _π/2

0

cos²(cosx)dx =

= Z π/2

0

cos²(sinx)dx =

= Z _π/2

0

(sin²(sinx) + cos²(sinx))dx =

= Z _π/2

0

1dx = π 2.

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Sums of the form 1 + ¹₂ +¹₃ +· · ·+¹_n often appear in mathematical formulas.

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Sums of the form 1 + ¹₂ +¹₃ +· · ·+¹_n often appear in mathematical formulas. It is beneficial to have a good estimate of such a sum, avoiding adding up the terms directly.

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x y

1 2 3 4 5 6 x

y

1 2 3 4 5 6

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x y

1 2 3 4 5 6 x

y

1 2 3 4 5 6 These two figures prove that

1 +1 2 +1

3+· · ·+1 n ≥

Z n+1 1

1

xdx = ln(n+ 1)−ln(1) = ln(n+ 1), 1

2+ 1

3+· · ·+ 1 n + 1

n+ 1 ≤ Z n+1

1

xdx = ln(n+ 1)−ln(1) = ln(n+ 1),

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x y

1 2 3 4 5 6 x

y

1 2 3 4 5 6 These two figures prove that

1 +1 2 +1

3+· · ·+1 n ≥

Z n+1 1

1

xdx = ln(n+ 1)−ln(1) = ln(n+ 1), 1

2+ 1

3+· · ·+ 1 n + 1

n+ 1 ≤ Z n+1

1

xdx = ln(n+ 1)−ln(1) = ln(n+ 1), so we have

ln(n+ 1)≤1 +1 2+ 1

3+· · ·+ 1

n ≤ln(n+ 1) + 1− 1 n+ 1.

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Indefinite integral / antiderivative. Methods: look up in the table of derivatives, simplify by u-substitution, simplify using integration by parts, combine the above methods.

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Definite integral: originally motivated by computing areas. Every continuous function, and even every piecewise continuous (bounded) function can be integrated.

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Main method for integration: use fundamental theorem of calculus (compute an antiderivative and subtract its values). Sometimes antiderivatives are not available, but a u-substitution applied to a part of the integral would help.

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Definite integral is defined via Riemann sums. Sometimes, handling a sum is easier if you interpret it as a Riemann sum, and examine the respective integral instead.

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Definite integral is defined via Riemann sums. Sometimes, handling a sum is easier if you interpret it as a Riemann sum, and examine the respective integral instead.

Next time: applications of the definite integral in geometry, science, and engineering.