### Video Transcript

A curve passes through the point zero, seven divided by 15 and the tangent at its point π₯, π¦ has slope eight π₯ multiplied by the square root of two π₯ plus one. What is the equation of the curve?

The question tells us that the curve passes through the point zero, seven fifteenths, and its tangent at the point π₯, π¦ has the slope eight π₯ multiplied by the square root of two π₯ plus one. We recall, if the slope has equation eight π₯ multiplied by the square root of two π₯ plus one, then the derivative of π¦ with respect to π₯ is equal to eight π₯ multiplied by the square root of two π₯ plus one. In particular, what this tells us is that the equation of the curve which we want to find, π¦, is an antiderivative of eight π₯ multiplied by the square root of two π₯ plus one. So, by taking the integral of eight π₯ multiplied by the square root of two π₯ plus one with respect to π₯, weβll find an equation of π¦ up to our constant of integration.

We see that our integrand is not in the standard form which we can integrate. So, weβre going to need to manipulate this. We might be tempted to use integration by parts. However, weβre going to do this by using the substitution π’ is equal to two π₯ plus one. So, by using the substitution π’ is equal to two π₯ plus one and differentiating both sides with respect to π₯, we have the derivative of π’ with respect to π₯ is equal to two. And although dπ’ by dπ₯ is not a fraction, it behaves a little bit like a fraction when using integration by substitution. This gives us the equivalent statement a half dπ’ is equal to dπ₯.

We see that our integrand contains a factor of eight π₯. So, weβre going to manipulate our expression of π’ is equal to two π₯ plus one to get an expression for eight π₯ in terms of π’. Multiplying both sides of this equation by four gives us that four π’ is equal to eight π₯ plus four. Then, we subtract four from both sides of the equation to get that four π’ minus four is equal to eight π₯. So, by using our substitution π’ is equal to two π₯ plus one, we have our integral is equal to the integral of four π’ minus four multiplied by the square root of π’ multiplied by a half with respect to π’.

Now, we can distribute over our parentheses and use the fact that the square root of π’ is equal to π’ to the power of a half to rearrange our integral to be the integral of two multiplied by π’ to the power of three over two minus two multiplied by π’ to the power of a half with respect to π’. We can now integrate this directly by recalling for constants π and π where π is not equal to negative one, to integrate π multiplied by π₯ to πth power with respect to π₯, we add one to the exponent and then divide by this new exponent plus our constant of integration π. To integrate our first term, we add one to the exponent to give us five over two. Then, instead of dividing by this new exponent, weβre going to multiply by the reciprocal.

Similarly, to integrate our second term, we add one to the exponent to give us three over two. Then, instead of dividing by our new exponent, weβre going to multiply by the reciprocal. Then, instead of adding a constant of integration for both of our terms, weβll add one constant of integration, which we will call π. At this point, we can notice both for first term and our second term share a factor of four π’ to the power of three over two. So, letβs factor this out. Factoring four multiplied by π’ to the power of three over two from our first term leaves us with π’ divided by five. And factoring four multiplied by π’ to the power of three over two from our second term leaves us with negative one-third.

Weβre now ready to rewrite our equation in terms of π₯ by substituting π’ is equal to two π₯ plus one. We can simplify this further by rewriting our fractions to have a common denominator. Doing this gives us six π₯ plus three over 15 minus five over 15 in our second set of parentheses. Then, we can evaluate this to give us six π₯ minus two divided by 15. Now, we see both terms in our numerator share a factor of two. So, letβs factor this out. This gives us that our curve π¦ is equal to eight multiplied by two π₯ plus one to the power of three over two multiplied by three π₯ minus one divided by 15 plus our constant of integration π.

The question tells us that the curve passes through the point zero, seven fifteenths. So, when π₯ is equal to zero, π¦ is equal to seven divided by 15. So, substituting π₯ is equal to zero and π¦ is equal to seven over 15 into our equation for the curve gives us seven fifteenths is equal to eight multiplied by two multiplied by zero. Plus one to the power of three over two multiplied by three multiplied by zero minus one over 15 plus our constant of integration π. We have that two multiplied by zero plus one is just equal to one, and then one to the power of three over two is also equal to one. Next, we have three multiplied by zero minus one all divided by 15 is just equal to negative one divided by 15.

So, this evaluates to give us that seven divided by 15 is equal to negative eight divided by 15 plus our constant of integration π. So, we add eight over 15 to both sides of our equation, and we see that our constant of integration π is equal to one. Therefore, after a small bit of rearrangement, we have shown that if a curve passes through the point zero, seven fifteenths and the tangent at the point π₯, π¦ has the slope eight π₯ multiplied by the square root of two π₯ plus one. Then, its equation is π¦ is equal to eight divided by 15 multiplied by two π₯ plus one to the power of three over two multiplied by three π₯ minus one plus one.