How do you evaluate the definite integral $\int (t^{3} - 9 t) d t$ $int (t^3-9t)dt$ from [-1,1]?

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Answer 1 · 2017-01-31T06:31:26

$0 .$ $0.$

Observe that the Integrand $f (t) = t^{3} - 9 t, t \in [- 1, 1],$ $f(t)=t^3-9t, t in [-1,1],$ is an odd
continuous function.

So, at a moment's glance, we can answer $0$ $0$ , if we know the following Result :

$int_-a^af(t)dt=0, if f : [-a,a] to RR" is continuous & odd on "[-a,a].$

Alternatively, $int_-1^1(t^3-9t)dt=[t^4/4-(9t^2)/2]_-1^1$

$=[(1/4-9/2)-(1/4-9/2)]=0.$

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How do you evaluate the definite integral int (t^3-9t)dt∫(t3−9t)dtint (t^3-9t)dt from [-1,1]?