Question

evaluate integral
integral 1 to2 3/t^4 dt

Answer 1

\[\int_{1}^{2}\frac{3}{t^4}dt\]What's confusing you about this one?

Answer 2

can you show me steps

Answer 3

im not sure how i would solve this...would i have to reciprocal?

Answer 4

\[\int_1^2{3t^{-4}}dt\]should be apparent now

Answer 5

ok...how you go about doing that??

Answer 6

first I take the constant 3 out\[3\int_1^2{t^{-3}dt}\]

Answer 7

wouldnt that becomes 1/3?

Answer 8

then I calculate the integral of \[t^{-3}dt\]before evaluating the definite integral:\[3\int_1^2{t^{-3}dt} = 3[\frac{t^{-2}}{-2}]_1^2 = -\frac{3}{2}[\frac{1}{t^2}]_1^2\]

Answer 9

\[\frac{a}{b}=ab^{-1}\]

Answer 10

after that, you substitute the limits. \[-\frac{3}{2}[\frac{1}{4}-\frac{1}{1}]\]

Answer 11

ok now i see i just dont understand the 3 constant taken out

Answer 12

Constants always get taken out of integrals (and differentials). For example, suppose I ask you to solve this:\[\int3dt.\]You can take the constant out:\[3\int dt,\]and still have the same expression:\[3t+C.\]

Answer 13

Also, it doesn't matter how tricky the constant looks:\[\int\frac{1}{3}dt=\frac{1}{3}\int dt,\]\[\int\pi dt=\pi\int dt.\]

Answer 14

ok so in this case it would be just 3