### Video Transcript

A group of laborers are digging a hole, where the rate of change of the volume π of the sand removed in cubic meters with respect to the time π‘ in hours is given by the relation dπ by dπ‘ is equal to π‘ plus 15. Calculate the volume of the sand dug out in five hours, rounded to the nearest hundredth.

In this question, weβre told that a group of laborers are a digging a hole. And weβre given a relation involving the volume of sand removed with respect to time. The derivative of this volume with respect to time is equal to π‘ plus 15, where the volume π of sand removed is given in the units cubic meters and the time π‘ is given in hours. We need to calculate the volume of sand removed in five hours. We need to round our answer to the nearest hundredth.

To answer this question, we start by noting the volume of sand removed is going to be given by the function π of π‘. Thatβs the amount of sand removed after π‘ hours. And therefore, since the question is asking us to find the amount of sand removed after five hours, the question wants us to evaluate π at five.

However, weβre not given the function π of π‘. Instead, weβre given dπ by dπ‘. Thatβs the rate of change in volume with respect to time. And since we know an expression for the derivative of π with respect to π‘, we need to find an antiderivative of this expression, something which differentiates to give us π‘ plus 15. And to do this, we can recall, using indefinite integration, we can find the most general antiderivative. In this case, π of π‘ will be the indefinite integral of its derivative dπ by dπ‘ with respect to π‘. And itβs worth reiterating this is only going to be true up to a constant of integration because weβre finding the most general antiderivative.

To evaluate this, letβs substitute the expression weβre given for dπ by dπ‘ into the equation. Weβre told dπ by dπ‘ is π‘ plus 15. Therefore, π of π‘ will be the indefinite integral of π‘ plus 15 with respect to π‘. We now need to evaluate this integral. Weβll do this term by term by using the power rule for integration, which tells us, for any real value of π not equal to negative one, the integral of π₯ to the πth power with respect to π₯ is equal to π₯ to the power of π plus one divided by π plus one plus a constant of integration πΆ. We add one to the exponent of our variable and divide by the new exponent.

Remember, weβre allowed to evaluate integrals term by term. So we can apply this to evaluate the integral of π‘. Well, itβs worth noting we can call the variable in our expression anything. The integral of π‘ to the πth power with respect to π‘ will be π‘ to the π plus one divided by π plus one plus πΆ. Therefore, by rewriting π‘ as π‘ to the first power, we add one to this exponent to get two and divide by the new exponent. The integral of π‘ to the first power is π‘ squared over two. And we can add a constant of integration now. However, weβll get a constant of integration when we integrate each term. And itβs easier to combine these into one constant of integration at the end of our expression.

Now, we need to evaluate the integral of the constant 15. And we can do this by using the power rule for integration. 15 is equal to 15 multiplied by π‘ to the zeroth power. However, itβs far easier to recall the derivative of a linear function is the coefficient of π‘. In other words, the derivative of 15π‘ with respect to π‘ is 15, which means that 15π‘ is an antiderivative of 15. Then, finally, we add our constant of integration πΆ. π of π‘ is equal to π‘ squared over two plus 15π‘ plus πΆ.

We want to find π evaluated at five. However, thereβs a problem. We still have this unknown constant of integration. And to find this value of πΆ, weβre going to need to know π of π‘ for some value of π‘. And we can do this directly from the question. Remember, π of π‘ is the amount of sand removed after π‘ hours. Therefore, if no time has passed, no sand will have been removed. π of zero is equal to zero. And since π‘ is equal to zero in this case, this is sometimes called the initial condition.

Now, we can substitute π‘ is equal to zero into our function π of π‘. π of zero is zero squared over two plus 15 times zero plus πΆ. And remember, this needs to be equal to zero since no sand has been removed at time π‘ is equal to zero. Zero squared over two is equal to zero, and 15 multiplied by zero is also equal to zero. Therefore, we just have πΆ is equal to zero.

We can then substitute πΆ is equal to zero into our function π of π‘. This gives us that π of π‘ is equal to π‘ squared over two plus 15π‘. And now we can substitute π‘ is equal to five into this expression to find the volume of sand removed after five hours. Substituting π‘ is equal to five into π of π‘ gives us π evaluated at five is five squared over two plus 15 multiplied by five, which we can evaluate is equal to 87.5.

But weβre not done yet. Remember, the question wants us to round our answer to the nearest hundredth. Therefore, we can add one extra decimal place of accuracy. And finally, remember, this value represents a physical amount. So we should give this units. Weβre told the volume π is measured in cubic meters provided π‘ is measured in hours. So we add the units of cubic meters to our answer, which gives us our final answer.

The volume of sand removed after five hours by the laborers to the nearest hundredth is 87.50 cubic meters.