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Evaluate x 2 sin x d x .

x 2 cos x + 2 x sin x + 2 cos x + C

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Integration by parts for definite integrals

Now that we have used integration by parts successfully to evaluate indefinite integrals , we turn our attention to definite integrals. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lower limits of integration.

Integration by parts for definite integrals

Let u = f ( x ) and v = g ( x ) be functions with continuous derivatives on [ a , b ] . Then

a b u d v = u v | a b a b v d u .

Finding the area of a region

Find the area of the region bounded above by the graph of y = tan −1 x and below by the x -axis over the interval [ 0 , 1 ] .

This region is shown in [link] . To find the area, we must evaluate 0 1 tan −1 x d x .

This figure is the graph of the inverse tangent function. It is an increasing function that passes through the origin. In the first quadrant there is a shaded region under the graph, above the x-axis. The shaded area is bounded to the right at x = 1.
To find the area of the shaded region, we have to use integration by parts.

For this integral, let’s choose u = tan −1 x and d v = d x , thereby making d u = 1 x 2 + 1 d x and v = x . After applying the integration-by-parts formula ( [link] ) we obtain

Area = x tan −1 x | 0 1 0 1 x x 2 + 1 d x .

Use u -substitution to obtain

0 1 x x 2 + 1 d x = 1 2 ln | x 2 + 1 | 0 1 .

Thus,

Area = x tan 1 x | 0 1 1 2 ln | x 2 + 1 | | 0 1 = π 4 1 2 ln 2.

At this point it might not be a bad idea to do a “reality check” on the reasonableness of our solution. Since π 4 1 2 ln 2 0.4388 , and from [link] we expect our area to be slightly less than 0.5, this solution appears to be reasonable.

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Finding a volume of revolution

Find the volume of the solid obtained by revolving the region bounded by the graph of f ( x ) = e x , the x -axis, the y -axis, and the line x = 1 about the y -axis.

The best option to solving this problem is to use the shell method. Begin by sketching the region to be revolved, along with a typical rectangle (see the following graph).

This figure is the graph of the function e^-x. It is an increasing function on the left side of the y-axis and decreasing on the right side of the y-axis. The curve also comes to a point on the y-axis at y=1. Under the curve there is a shaded rectangle in the first quadrant. There is also a cylinder under the graph, formed by revolving the rectangle around the y-axis.
We can use the shell method to find a volume of revolution.

To find the volume using shells, we must evaluate 2 π 0 1 x e x d x . To do this, let u = x and d v = e x . These choices lead to d u = d x and v = e x = e x . Substituting into [link] , we obtain

Volume = 2 π 0 1 x e x d x = 2 π ( x e x | 0 1 + 0 1 e x d x ) Use integration by parts . = −2 π x e x | 0 1 2 π e x | 0 1 Evaluate 0 1 e x d x = e x | 0 1 . = 2 π 4 π e . Evaluate and simplify .
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Evaluate 0 π / 2 x cos x d x .

π 2 1

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Key concepts

  • The integration-by-parts formula allows the exchange of one integral for another, possibly easier, integral.
  • Integration by parts applies to both definite and indefinite integrals.

Key equations

  • Integration by parts formula
    u d v = u v v d u
  • Integration by parts for definite integrals
    a b u d v = u v | a b a b v d u

In using the technique of integration by parts, you must carefully choose which expression is u. For each of the following problems, use the guidelines in this section to choose u. Do not evaluate the integrals.

x 3 e 2 x d x

u = x 3

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y 3 cos y d x

u = y 3

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e 3 x sin ( 2 x ) d x

u = sin ( 2 x )

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Find the integral by using the simplest method. Not all problems require integration by parts.

ln x d x ( Hint: ln x d x is equivalent to 1 · ln ( x ) d x . )

x + x ln x + C

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tan −1 x d x

x tan −1 x 1 2 ln ( 1 + x 2 ) + C

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x sin ( 2 x ) d x

1 2 x cos ( 2 x ) + 1 4 sin ( 2 x ) + C

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x e x d x

e x ( −1 x ) + C

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x 2 cos x d x

2 x cos x + ( −2 + x 2 ) sin x + C

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ln ( 2 x + 1 ) d x

1 2 ( 1 + 2 x ) ( −1 + ln ( 1 + 2 x ) ) + C

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e x sin x d x

1 2 e x ( cos x + sin x ) + C

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x e x 2 d x

e x 2 2 + C

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sin ( ln ( 2 x ) ) d x

1 2 x cos [ ln ( 2 x ) ] + 1 2 x sin [ ln ( 2 x ) ] + C

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Practice Key Terms 1

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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