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

Answer 1

#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#.
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Answer 2

To find the definite integral of ( t^3(1 + t^4)^3 ) with respect to ( t ) from ( -1 ) to ( 1 ), you can use integration techniques. One common approach is to first expand ( (1 + t^4)^3 ) using the binomial theorem, then integrate each term separately, and finally evaluate the result at the upper and lower limits of integration. Alternatively, you can use a computer algebra system or integral tables to compute the integral directly.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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