How do you find the definite integral of #t^3(1 + t^4)^3 dt# from #[-1, 1]#?

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Answer 1 · 2017-08-15T01:17:02

#0#

The quick way

This is an odd function with domain #RR#.

That is: #f(-t) = -f(t)# for all #t# in #RR#

For any odd function #f# whose domain includes #[-a,a]#,

#int_-a^a f(x) dx = 0#

The long way

Use substitution with #u = 1+t^4#, so #du = 4t^3 dt#

And #t=-1# #rArr# #u = 2# and #t=1# #rArr# #u = 2#.

So the integral becomes

#int_2^2 1/4u^3 du# = 0#.

Because #int_a^a f(x) dx = 0#.

The very long way

Finish integrating

#int t^3(1+t^4)^3 dt = 1/4 int u^4 du#

# = u^5/20#

# = (1+t^4)^5/20#

Now evaluate from #-1# to #1# to get #0#.