evaluation of definite integral (1 Viewer)

braintic · Apr 21, 2014

juantheron said:
$\displaystyle I = \int_{0}^{2}\left(3x^2-3x+1\right)\cdot \cos\left(x^3-3x^2+4x-2\right)dx$

Are you asking how to do this, or are you setting it for people to do?

Carrotsticks · Apr 21, 2014

He is setting it for people to do, that's what he does around here.

Sy123 · Apr 21, 2014

juantheron said:
$\displaystyle I = \int_{0}^{2}\left(3x^2-3x+1\right)\cdot \cos\left(x^3-3x^2+4x-2\right)dx$

$I = \int_0^2 (3x(x-1)+1) \cos((x-1)((x-1)^2+1)) \ dx$

$\\ u = x-1 \\ \Rightarrow I = \int_{-1}^1 (3u(u+1)+1) \cos(u(u^2+1)) \ du \\ \\ u = -z \\ \Rightarrow I = \int_{-1}^1 (-3z(1-z) + 1) \cos(z(z^2+1)) \ dz \\ $Adding side by side$ \\ \therefore \ 2I = \int_{-1}^1 (3x(x+1) + 1 + (-3x)(1-x) + 1) \cos(x(x^2+1)) \ dx \\ I = \int_{-1}^1 (3x^2+1) \cos(x(x^2+1)) \ dx = \sin(x(x^2+1))|_{-1}^1 = 2\sin(2)$

If you wish to post problems for the community to solve with the intention of helping others practise, then I encourage you to post them in the MX2 Integration Marathons for all definite and indefinite integrals, in the HSC 4U Marathon - Advanced Level if the question is not the average 4U type question, otherwise post it in the HSC 4U Marathon.

evaluation of definite integral (1 Viewer)

juantheron

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braintic

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Carrotsticks

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Sy123

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