Video Transcript
If π double prime of π₯ equals
three π₯ to the power of five plus three π₯ cubed plus five π₯ plus two, determine
π of π₯. We know how to find the derivative
of different functions. For example, if we take a
polynomial π of π₯ equals two π₯ cubed plus three π₯ squared plus five π₯ minus
one, we can differentiate this term by term using the power rule which tells us to
multiply the coefficient by the power and then reduce the power by one. We also know that constants
differentiate to zero.
So if we apply this to each term,
we find that π prime of π₯ equals six π₯ squared plus six π₯ plus five. We can differentiate this again to
find π double prime of π₯. This gives us 12π₯ plus six. But here for this problem, weβre
wanting to find the antiderivative, βanti-β meaning opposite. So weβve been given π double prime
of π₯ and weβve been asked to determine π of π₯. So we need to work backwards. But how do we do this? Well, if we know how to
differentiate, then we can definitely do the opposite, its integration. So we take our rules for
differentiating and we reverse them.
The opposite of reducing the power
by one is adding one to the power. The opposite of multiplying the
coefficient by the power is dividing by the power. Note that here we multiplied by the
old power. So for the reverse, we divide by
the new power. And because constants differentiate
to zero, we need to add a constant of integration. So letβs now find the
antiderivative of π double prime of π₯. Just like how we differentiate term
by term, we can integrate term by term. We start with three π₯ to the power
of five. And we increase the power by one to
give us three π₯ to the power of six. And then we divide by the new power
of six.
We can actually simplify this term
as three over six is one over two. We follow the same rules for our
next term. We add one to the power to get
three π₯ to the power of four and divide by the new power of four. Similarly, we increase the power of
five π₯ to get five π₯ squared and divide by the new power. And now we integrate the constant,
two. Remember how, in the first example,
we looked at five π₯ differentiated to give us five. So five integrates to give us five
π₯. In the same way two integrates to
give us two π₯. Weβre not quite finished with this
step because weβve got to add a constant of integration, πΆ. So that gives us π prime of π₯ but
weβre not finished yet because weβre trying to determine π of π₯. So we must integrate one more
time.
Following the same procedure as
before, π₯ to the power of six over two integrates to π₯ to the power of seven over
two divided by seven. We could write this as π₯ to the
power of seven over two multiplied by one over seven, which is just π₯ to the power
of seven over 14. And now we move on to our next
term. Adding one to the power and
dividing by the new power gives us three π₯ to the power of five over four all
divided by five, which we can rewrite as three π₯ to the power of five over 20.
Now, we move on and integrate five
π₯ squared over two and then we simplify to five π₯ cubed over six. And then two π₯ integrates to two
π₯ squared over two. And the twos cancel to give us π₯
squared. And now we also have to integrate
our constant πΆ. In the same way that two integrated
to two π₯, πΆ integrates to πΆπ₯. And finally, we need to add a
constant of integration, which this time weβll call π·. So that gives us our final
answer. π of π₯ equals π₯ to the power of
seven over 14 plus three π₯ to the power of five over 20 plus 5π₯ cubed over six
plus π₯ squared plus πΆπ₯ plus π·.
