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### Section 5.2 : Computing Indefinite Integrals

24. Determine $$h\left( t \right)$$ given that $$h''\left( t \right) = 24{t^2} - 48t + 2$$, $$h\left( 1 \right) = - 9$$ and $$h\left( { - 2} \right) = - 4$$.

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Hint : We know how to find $$h\left( t \right)$$ from $$h'\left( t \right)$$ but we don’t have that. We should however be able to determine the general formula for $$h'\left( t \right)$$ from $$h''\left( t \right)$$ which we are given.
Start Solution

Because we know that the 2nd derivative is just the derivative of the 1st derivative we know that,

$h'\left( t \right) = \int{{h''\left( t \right)\,\,dt}}$

and so to arrive at a general formula for $$h'\left( t \right)$$ all we need to do is integrate the 2nd derivative that we’ve been given in the problem statement.

$h'\left( t \right) = \int{{24{t^2} - 48t + 2\,\,dt}} = 8{t^3} - 24{t^2} + 2t + c$

Don’t forget the “+c”!

Hint : From the previous two problems you should be able to determine a general formula for $$h\left( t \right)$$. Just don’t forget that $$c$$ is just a constant!
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Now, just as we did in the previous two problems, all that we need to do is integrate the 1st derivative (which we found in the first step) to determine a general formula for $$h\left( t \right)$$.

$h\left( t \right) = \int{{8{t^3} - 24{t^2} + 2t + c\,\,dt}} = 2{t^4} - 8{t^3} + {t^2} + ct + d$

Don’t forget that $$c$$ is just a constant and so it will integrate just like we were integrating 2 or 4 or any other number. Also, the constant of integration from this step is liable to be different that the constant of integration from the first step and so we’ll need to make sure to call it something different, $$d$$ in this case.

Hint : To determine the value of the constants of integration, $$c$$ and $$d$$, we have the value of the function at two values that should help with that.
Show Step 3

Now, we know the value of the function at two values of $$z$$. So let’s plug both of these into the general formula for $$h\left( t \right)$$ that we found in the previous step to get,

$\begin{array}{*{20}{c}}{ - 9 = h\left( 1 \right) = - 5 + c + d}\\{ - 4 = h\left( { - 2} \right) = 100 - 2c + d}\end{array}$

Solving this system of equations (you do remember your Algebra class right?) for $$c$$ and $$d$$ gives,

$c = \frac{{100}}{3}\hspace{0.5in}d = - \frac{{112}}{3}$

The function is then,

$\require{bbox} \bbox[2pt,border:1px solid black]{{h\left( t \right) = 2{t^4} - 8{t^3} + {t^2} + \frac{{100}}{3}t - \frac{{112}}{3}}}$