Compact Disc Sales Sales of the latest recording by Brittania, a British rock group, are current

Compact Disc Sales Sales of the latest recording by...

Key Concepts

Rate Functions and Accumulation

This concept involves understanding that the rate function represents the instantaneous change of a quantity, and the total amount accumulated over time can be found by integrating the rate function. In problems involving sales or any continuously changing quantity, the integral from the starting time to a given time yields the total quantity accumulated over that period.

Integration Techniques for Exponential Functions

When faced with an integral that includes an exponential function multiplied by a polynomial term, standard integration techniques are required. In many cases, integration by parts is used to handle the polynomial-exponential product. Recognizing how to decompose the integrand into parts that simplify after differentiation and integration is crucial for obtaining the correct antiderivative.

Transcript

00:02 In the question given cells value in terms of function of t that is given as 2 t times of e to the power minus 0 .1 t units per week where t is time in terms of which in units is weak and cells is in term of unit and unit will show 10 ,000 cds.

00:42 Now in the question it is asking to find the total number of cds.

00:48 So we have to find the total number of cd by using the given function.

00:59 So it will be written as total number of cds in term of t as a function that is 0 to t to t times of e to the power minus 0 .1 d t so if this function will show the total number of cd so as you see i have to find this definite integral value so for finding this first let assume that u is equal to t so i can say that du will become that is d t and again let assume that dv will be equal to two times of e to the power minus 0 .1 t d t so we will become that is after integrating value is minus 20 times of e to the power minus 0 .1 t so this is the value of v now my next step is i am applying the formula of integration by part which is written as integral udv will be equal to uv minus integral v d u so this is a formula of indication by part so in this formula i will substituting all the data so again it will be written as that is integral udb means u is as you know above see that i write u is equal to t times of dv means two times of e to the power of minus 0 .1 t d t will be equal to uv uv means t times of minus 20 times of 20 times of e to the power minus 0 .1 t minus integral integral v will be that is minus 20 times of e to the power minus 0 .1 t times of du du is that is d t so solve this again then i will say that integral t times of two times of e to the power minus 0 .1 dt will be equal to that that is minus 20 t e to the power minus 0 .1 t minus minus become plus and in the bracket 20 will come outside because it is a constant value now we have to find the value of integral e to the power minus 0 .1 d t so value will be that is minus 10 times of e to the power minus 0 .1 d t d t so value will be that is minus 10 times of e to the power minus 0 .1 t plus c now simplify this then i will get integral 2 t e to the power minus 0 .1 t d t is equal to minus 20 t times of e to the power minus 0 .1 t minus 200 times of e to the power now as you see in the question in the beginning i said i had said that the total number of cd in function will be equal to 0 to t integral 2 t e to the power minus 0 .1 t d t so as you see in the result i had find only the value of integral 2 t e to the power minus 0 .1 t d t so as you see in the result i had find only the value of integral 2 t e to the power minus 0 .1 t d t d t so now i will apply the limits that is 0 to t so i can find the so after applying this limit i will get the value of total number of cds as a function in term of t that is will be written as 0 to t 2 t e to the power minus 0 .1 times of d t e to the power minus 0 .1 t d t will be equal to.

06:06 Here i will forget to write the term t on above and this step.

06:12 So i had write t...

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