Video Transcript
Determine the integral of four ππ to the power of three π₯.
To integrate four ππ to the power of three π₯, we actually have a rule. And the rule weβre gonna use tells us that the integral of π to the power of ππ₯ plus π is equal to one over π π to the power of ππ₯ plus π plus π. And this is true when π is not equal to zero. And this comes from something that youβd have recovered previously which is that if you differentiate π to the power of ππ₯ plus π, then the result is ππ to the power of ππ₯ plus π. And that relationship was found using the chain rule.
Okay, so weβve now got a general rule for integrating π to the power of ππ₯ plus π which we have derived from the derivative of π to the power of ππ₯ plus π. Things to watch out for here, donβt forget the π because obviously, we need to add π as this is our constant of integration. And just to remind you why we need that constant of integration, weβre gonna look at π¦ equals four π₯ squared plus three. If I actually differentiate four π₯ squared plus three, then the derivative is gonna be equal to eight π₯. However, if I wanted to use integration to actually find the original function, so what π¦ was equal to, then what I would have to do is integrate eight π₯. And then, what I get if I use the rule for integration would be eight π₯ to the power of one plus one, cause you add one to the exponent, and then divided by the new exponent, which would give me eight π₯ squared over two, which would give me four π₯ squared.
So if we look back at our original function, okay great. Yep, weβve got our four π₯ squared. And actually, we can see that we havenβt got the positive three. So therefore, we have to add π, so add our constant of integration. And thatβs because if weβre actually working to find our function, so find π¦, we donβt know if there was an additional number added on to the original function. So if weβd had four π₯ squared plus nine and four π₯ squared minus 12, that still will have differentiated to eight π₯. So thatβs why we need the constant of integration, so that we could say that we know something could be added to the term that weβve already got.
Okay, so now we know what we need to do. Letβs use it to actually determine the integral of our expression. So first, well weβre gonna take a look. Weβve actually got four π in front of our π to the power of three π₯. Well, this doesnβt actually change the integration. So this coefficient wonβt affect the integration. So we can carry on and integrate. So if we look at the example weβve got, so we look at our rule, we can see that our π is gonna be equal to three. Okay, so now letβs use that rule and integrate our expression.
So when we have actually differentiated four ππ to the power of three π₯, weβre gonna get four π over three π to the power of three π₯ and then plus π. And thatβs because if we look back at our general rule, we can see that itβs gonna be one over π. Well, in this case, itβs gonna be the coefficient, which is four π, over our π, which is three. And then the π to the power of three π₯ remains unchanged. And then we have to remember to add the constant of integration. So we get our final answer that if we integrate four ππ to the power of three π₯, the result is four π over three π to the power of three π₯ plus π.