Let

Upper I Subscript nInequals=Integral from nothing to nothing x Superscript n Baseline e Superscript negative x squared Baseline font size decreased by 6 dx∫xne−x2 dx​,

where n is a nonnegative integer. Complete parts​ (a) through​ (e) below.

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Part 1

a.

Upper I 0I0equals=Integral from nothing to nothing e Superscript negative x squared Baseline font size decreased by 6 dx∫e−x2 dx

cannot be expressed in terms of elementary functions. Evaluate

Upper I 1I1.

Upper I 1I1equals=enter your response here

Part 2

b. Use integration by parts to evaluate

Upper I 3I3.

Upper I 3I3equals=enter your response here

Part 3

c. Use integration by parts and the result of part​ (b) to find

Upper I 5I5.

Upper I 5I5equals=enter your response here

Part 4

d. Show​ that, in​ general, if n is​ odd, then

Upper I Subscript nInequals=negative one half e Superscript negative x squared−12e−x2p Subscript n minus 1pn−1​(x),

where

p Subscript n minus 1pn−1

is an even polynomial of degree

nminus−1.

Choose the correct answer below.

A.

Upper I Subscript 2 n plus 1I2n+1equals=negative one half e Superscript negative x squared Baseline x Superscript 2 n−12e−x2x2nplus+nUpper I Subscript 2 n minus 1I2n−1

B.

Upper I Subscript 2 n plus 1I2n+1equals=negative one half e Superscript negative x squared Baseline x Superscript 2 n−12e−x2x2nplus+nUpper I Subscript 2 nI2n

C.

Upper I Subscript 2 n plus 1I2n+1equals=negative one half e Superscript negative x squared Baseline x Superscript 2 n−12e−x2x2nminus−nUpper I Subscript 2 n minus 1I2n−1

D.

Upper I Subscript 2 n plus 1I2n+1equals=negative one half e Superscript negative x squared Baseline x Superscript 2 n−12e−x2x2nminus−nUpper I Subscript 2 nI2n

Part 5

e. Argue that if n is​ even, then

Upper I Subscript nIn

cannot be expressed in terms of elementary functions.

A.

If n is​ even, and a​ u-substitution is made for

x squaredx2​,

an integral of the form

Integral from nothing to nothing x ln u font size decreased by 6 du∫xlnu du

will always​ remain, even after all other integration techniques have been performed. The solution to

Integral from nothing to nothing x ln u font size decreased by 6 du∫xlnu du

cannot be expressed in terms of elementary functions.

B.

If n is​ even, and a​ u-substitution is made for

x squaredx2​,

an integral of the form

Integral from nothing to nothing x e Superscript negative u Baseline font size decreased by 6 du∫xe−u du

will always​ remain, even after all other integration techniques have been performed. The solution to

Integral from nothing to nothing x e Superscript negative u Baseline font size decreased by 6 du∫xe−u du

cannot be expressed in terms of elementary functions.

C.

If n is​ even, the number of steps required in integrating by parts becomes infinite. There is no way to express this result in terms of elementary functions.