$=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k$

The first integral is trivial since $(\tan t)'=\sec^2t$ .

The second can be done by substituting $u=t^2-1$ :

$u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C$

The third can be found by integrating by parts:

$u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t$

$\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8$

$\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C$

8 0

4 years ago

Becca has a balance of $22.90 at the beginning of May for clothes. Her allotment for the month is $50.00. She buys a two new out

a_sh-v [17]

Add the balance and allotment:

22.90 + 50.00 = 72.90

This is how much she can actually spend.

She spent 85.50.

Subtract the two:

85.50 - 72.90 = 12.60

Since she spent more than she actually could be balance would be a negative number.

The balance would be - $12.60

4 0

4 years ago

Determine whether the graph of the equation is vertical, horizontal, or neither.

julia-pushkina [17]

Answer:

horizontal

Step-by-step explanation:

7 0

3 years ago