Week #13 - Integration by Parts & Numerical Integration ...

Week #13 - Integration by Parts & Numerical Integration

Section 7.2

From "Calculus, Single Variable" by Hughes-Hallett, Gleason, McCallum et. al. Copyright 2005 by John Wiley & Sons, Inc.

This material is used by permission of John Wiley & Sons, Inc.

SUGGESTED PROBLEMS

2. Integrate by parts,

selecting: u = t dv = sin t dt

so

du = dt v = - cos t

Using the integration by parts formula, udv = uv - vdu,

t sin t dt = -t cos t - (- cos t) dt = -t cos t + sin t + C

3. Integrate by parts,

selecting: u = t2 dv = sin t dt

so

du = 2t dt v = - cos t

Using the integration by parts formula,

t2 sin t dt = -t2 cos t - (-2t cos t) dt = -t2 cos t + 2 t cos t dt

This new integral is simpler (the power of t went from squared to linear), and if we integrate by parts again, we should get a solvable integral:

selecting: u = t dv = cos t dt

so

du = dt v = sin t

2 t cos t dt = 2 t sin t - sin t dt = 2 [t sin t + cos t] + C Combining this with the previous result, the final integral evaluates to

t2 sin t dt = -t2 cos t + 2t sin t + 2 cos t + C

9. To integrate by parts, we must select u = ln x, because we don't know how to integrate

it:

selecting: u = ln x dv = x3 dx

so

du

=

1 x

dx

v

=

x4 4

Using the integration by parts formula,

x3

ln x

dx

=

x4 4

ln x

-

x4 4

1 x

dx

=

x4 4

ln x -

1 4

x3 dx

=

x4 4

ln x

-

1 16

x4

+

C

1

17. Since everything seems to be in terms of ( + 1), it might help to do a substitution first:

Let w = ( + 1, so dw = d, and ( + 1) sin( + 1) d = w sin(w)dw

This we can evaluate by integration by parts,

selecting: u = w dv = sin w dw

so

du = dw v = - cos w

Using the integration by parts formula,

w sin(w)dw = -w cos w - (- cos w)dw = -w cos w + sin w = -( + 1) cos( + 1) + sin( + 1) + C

21. Neither substitution nor integration by parts looks to helpful at first glance. When possible, it can help to separate sums in the numerator to obtain two simpler integrals:

t+7 5 - t

dt

=

t 5 -

t

dt

+

7

I1

1 5 -

t

dt

I2

The second integral, I2, can be done by substitution, or by guess and check:

7

1 5 -

t

dt

=

7

(5 - t)-1/2 = -14(5 - t)1/2 + C

The first integral, I1, is more complicated, but with what we just found, we should be able to use integration by parts:

I1 =

t 5 -

t

dt

=

t(5 - t)-1/2 dt

selecting: u = t dv = (5 - t)-1/2 dt

so

du = dt v = -2(5 - t)1/2

Using the integration by parts formula,

I1 = t(5 - t)-1/2 dt = -2t(5 - t)1/2 - (-2(5 - t)1/2) dt

=

-2t(5

-

t)1/2

-

2

2 3

(5

-

t)3/2

+

C

Combining I1 and I2, and combining the constants into C2, we obtain

t+7 5 - t

dt

=

I1

+

I2

=

-2t(5

-

t)1/2

-

4 3

(5

-

t)3/2

-14(5

-

t)1/2

+C2

I1

I2

2

25. This looks like a classical substitution problem, given the x2 inside the arctan, and its derivative, x out front. Let's start with that substitution:

Let w = x2, so dw = 2x dx, so

x arctan x2

dx

=

1 2

arctan w dw

However, we only know the derivative of arctan, not its integral, so we need to use integration by parts,

selecting: u = arctan w dv = dw

so

du

=

1 1+w2

v=w

Using the integration by parts formula,

1 2

arctan w

dw

=

1 2

(w

arctan

w

-

w 1 + w2

dw)

This new integral can also be solved by substitution, since the derivative of the denominator will involve a w term.

Let z = 1 + w2, so dz = 2wdw:

w 1 + w2

dw

=

1 2

1 dz

z

=

1 2

ln z

=

1 2

ln(1

+

w2)

Combining all the parts, and then substituting back into the original x variable,

x arctan x2

dx

=

1 2

(w

arctan w

-

1

w + w2

dw)

=

w

arctan w 2

-

1 4

ln(1

+

w2)

+

C

=

x2

arctan x2 2

-

1 4

ln(1

+

x4) +

C

35. We only know how to differentiate ln, so trying integration by parts makes sense,

selecting: u = ln(1 + t) dv = dt

so

du

=

1 1+t

dt

v=t

Using the integration by parts formula,

5

ln(1 + t) dt = t ln(1 + t) -

0

t 1 + t dt

3

The new integral can be easily solved with the substitution w = 1 + t, so t = w - 1 and dw = dt:

t 1+t

dt

=

w

- w

1

dw

=

w w dw -

1 w dw

= w - ln w + C = 1 + t - ln(1 + t) + C

so

With the limits,

ln(1 + t) dt = t ln(1 + t) -

t 1 + t dt

= t ln(1 + t) - (1 + t - ln(1 + t) + C)

= (t + 1) ln(1 + t) - 1 - t + C2 if we let -C = C2

5

5

ln(1 + t) dt = (t + 1) ln(1 + t) - 1 - t

0

0

= (6 ln 6 - 6) - (ln(1) - 1) = 6 ln 6 - 5 5.751

QUIZ PREPARATION PROBLEMS

38. (a) This integral can be evaluated using integration by parts with u = x, dv = sin x dx. (b) We evaluate this integral using the substitution w = 1 + x3. (c) We evaluate this integral using the substitution w = x2. (d) We evaluate this integral using the substitution w = x3. (e) We evaluate this integral using the substitution w = 3x + 1. (f) This integral can be evaluated using integration by parts with u = x2, dv = sin x dx. (g) This integral can be evaluated using integration by parts with u = ln x, dv = dx.

39. You should have some sense of a sketch of this graph: it will look like the sine graph, but the amplitude will be growing with linearly x because the sin(x) is multiplied by x. Because it is a product of x and sin(x), the function will still have y = 0 values, crossing the x axis, at the same points when sin(x) = 0. The first arch of the curve is therefore between x = 0 and x = , as these are the first two points when f (x) = 0.

This means the area we're looking for will be represented by the integral x sin(x) dx.

0

To evaluate this, we'll use integration by parts:

selecting: u = x dv = sin x dx

so

du = dx v = - cos x

x sin(x) dx = x(- cos x) - - cos(x) dx

0

0

0

= -x cos(x) + sin(x)

0

= [(- cos() + sin()] - [(-0 cos(0) + sin(0)]

= -(-1) =

4

47.

We integrate by parts.

Let u = xn and dv = cos ax dx, so du = nxn-1 dx and v =

1 a

sin

ax.

Then

xn

cos axdx

=

1 xn sin ax a

-

(nxn-1)( 1 sin ax) dx a

=

1 xn sin ax a

-

n a

xn-1 sin ax dx.

51. Since f (x) = 2x, integration by parts tells us that

10

10

10

f (x)g(x)dx = f (x)g(x) - f (x)g(x)dx

0

0

0

10

= f (10)g(10) - f (0)g(0) - 2 xg(x)dx.

0

10

We can use let and right Riemann Sums with x = 2 to approximate xg(x)dx:

0

Left sum 0 ? g(0)x + 2 ? g(2)x + 4 ? g(4)x + 6 ? g(6)x + 8 ? g(8)x = (0(2.3) + 2(3.1) + 4(4.1) + 6(5.5) + 8(5.9))2 = 205.6

Right sum 2 ? g(2)x + 4 ? g(4)x + 6 ? g(6)x + 8 ? g(8)x + 10 ? g(10)x = 2(3.1) + 4(4.1) + 6(5.5) + 8(5.9) + 10(6.1))2 = 327.6.

A good estimate for the integral is the average of the left and right sums, so

10 0

xg(x)dx

205.6

+ 2

327.6

=

266.6.

Substituting values for f and g, we have

10

10

f (x)g(x)dx = f (10)g(10) - f (0)g(0) - 2 xg(x)dx

0

0

102(6.1) - 02(2.3) - 2(266.6) = 76.8 77.

5

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