### Video Transcript

Find the equation of the curve that
passes through the point negative two, one given that the gradient of the tangent to
the curve is negative 11π₯ squared.

In this question, we want to find
the equation of a curve. And to do this, weβre given some
information about our curve. Weβre told that our curve passes
through the point negative two, one and the gradient of the tangent to this curve is
equal to the function negative 11π₯ squared. So letβs break down the two pieces
of information weβre given about our curve. First, if we wanted to write the
equation of this curve in the form π¦ is equal to π of π₯, then saying that our
curve passes through the point negative two, one means when π₯ is equal to negative
two, π¦ should be equal to one.

So we can substitute these values
into the equation for our curve. We should have that one is equal to
π evaluated at negative two. Next, weβre told the gradient of
the tangent to our curve is negative 11π₯ squared. And remember, we know the gradient
of the tangent to our curve at a value of π₯ is as rate of change of π¦ with respect
to π₯. In other words, this is telling us
dπ¦ by dπ₯ is equal to negative 11π₯ squared. So we need to find the equation of
a curve given its slope and a point on the curve. And to do this, weβre going to need
to find an antiderivative of negative 11π₯ squared. And we know how to find
antiderivatives by using integration. The integral of negative 11π₯
squared with respect to π₯ will give us the most general antiderivative of negative
11π₯ squared.

And remember, the reason weβre
finding this is we know the derivative of our expression for π¦ with respect to π₯
has to be equal to negative 11π₯ squared. So we need to integrate negative
11π₯ squared with respect to π₯. We can do this by using the power
rule for integration. We recall this tells us for any
real constants π and π, where π is not equal to negative one, the integral of π
times π₯ to the πth power with respect to π₯ is equal to π times π₯ to the power
of π plus one divided by π plus one plus a constant of integration πΆ. We add one to our exponent and then
divide by this new exponent.

In our case, the value of π is
negative 11 and the value of π is two. So we add one onto our exponent of
two to get an exponent of three and then divide by this new exponent. We get negative 11π₯ cubed over
three plus a constant of integration πΆ. And of course, the derivative of
this expression with respect to π₯ is equal to negative 11π₯ squared. So far, for our curve π¦ is equal
to negative 11π₯ cubed over three plus a constant of integration πΆ, we have dπ¦ by
dπ₯ is equal to negative 11π₯ squared. However, remember, we still need
this curve to pass through the point negative two, one. And to do this, we need to
substitute π₯ is equal to negative two and π¦ is equal to one in to our curve.

Substituting π₯ is equal to
negative two and π¦ is equal to one into the equation for our curve, we get one is
equal to negative 11 times negative two cubed all over three plus our constant of
integration πΆ. Now all we need to do is solve this
equation for πΆ. Rearranging this equation, we see
that πΆ should be equal to negative 85 divided by three. Now all we need to do is use this
value of πΆ in the equation for our curve, and this gives us our final answer. Therefore, we were able to show the
equation of the curve that passes through the point negative two, one and which has
the gradient of its tangent line at π₯ given by negative 11π₯ squared has the
equation π¦ is equal to negative 11 over three times π₯ cubed minus 85 over
three.