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evaluate the integral a=1 b=2 f(x)= 3/t^4
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thing you mean f(t) = 3/t^4
\[\int\limits_{1}^{2} \frac{3}{t^4} dt \]
\[= 3 \int\limits_{1}^{2} t^{-4} dt\]
\[= 3( \frac{t^{-3}}{-3} ) \]
between the limits
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\[- [ \frac{1}{t^3}] \] between 2 and 1
\[= - ( [ 1/8] - [1/4] ) = (1/8)\]
1/8
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