Question

Functions defined by integrals, please let me know if you have any questions regarding the materials, I'd be more than happy to help. Thanks!

Answer 1

Given the function

`D\left(t\right)=(6t)/(1+2t)`

Notice that

`\begin{gathered} D\left(t\right)=6\left((t)/(1+2t)\right)=6\left((1)/(2)-(1)/(4t+2)\right) \\ \Rightarrow D\left(t\right)=6\left((1)/(2)-(1)/(2\left(2t+1\right))\right) \end{gathered}`

Then, integrate D(t) from t=0 to t=3, as shown below

`\Rightarrow\int_0^3D\left(t\right)dt=\int_0^36\left((1)/(2)-(1)/(2\left(2t+1\right))\right)dt=6\left((1)/(2)\int_0^3dt-(1)/(2)\int_0^3(dt)/(\left(2t+1\right))\right?`

Then,

`\Rightarrow\int_0^3D\left(t\right)dt=6\left((1)/(2)*3-(1)/(2)\int_0^3(dt)/(2t+1)\right)`

As for the remaining integral, set x=2t+1; then dx=2dt. Solving using that substitution,

as for the integration limits, when t=3, x=7, and when t=0, x=1.

`\int_0^3(dt)/(2t+1)=(1)/(2)\int_1^7(dx)/(x)=(1)/(2)\left(ln\left(7\right)-ln\left(1\right)\right?=(1)/(2)\left(ln\left((7)/(1)\right)\right)=(1)/(2)ln\left(7\right)`

Therefore,

`\Rightarrow\int_0^3D\left(t\right)dt=6\left((3)/(2)-(1)/(2)\left((1)/(2)ln\left(7\right)\right)\right)=9-(3)/(2)ln\left(7\right)`

Rounding to three decimal places,

`\Rightarrow\int_0^3D\lparen t)dt\approx6.081`

Thus, the answer is 6.081