Basic Integration: Exercise 2

by Batool Akmal

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    00:01 Let's have a look at our second integral.

    00:04 You can see here that this looks like a function of a function equation.

    00:09 So let's see how we deal with this.

    00:11 Remember, we've done a very similar example in the previous lecture.

    00:16 So let's attempt to integrate this.

    00:18 It's still an indefinite integral, so there are no limits.

    00:21 That means we will put a plus C at the end and it's to the power of three dx.

    00:27 So remember what we said.

    00:28 If you're integrating something like this, we're trying to do the exact opposite of the chain rule when we differentiate.

    00:34 So, this is the outside function, this is the inside function.

    00:40 To start off, we're going to integrate the outside function without changing anything on the inside.

    00:46 So to integrate, you add one to the power, divide by new power.

    00:50 And then rather than multiplying with the differential of the inside, the differential of the inside in this case is 10.

    00:57 We are going to divide it by the differential of the inside, and obviously those two values at the bottom multiply.

    01:03 So we are adding one to the power, dividing by new power, dividing by the differential of the inside.

    01:10 Put plus C there or later.

    01:12 So we've got 10x minus 5 to the power of 4 all over 40 plus C.

    01:20 And that's your integral.

    About the Lecture

    The lecture Basic Integration: Exercise 2 by Batool Akmal is from the course Basic Integration.

    Included Quiz Questions

    1. [(3x+1)³ / 9]+c
    2. [(3x+1)³ / 30]+c
    3. [(3x+1)³ / 3]+c
    4. [(3x+1)³ / 9]
    5. [(3x+1)³ / 90]+c
    1. [(10x-4)²³ / 230]+c
    2. [(10x-4)²³ / 23]+c
    3. [(10x-4)²³ / 230]
    4. [(10x+4)²³ / 230]+c
    5. [(10x-4)²² / 230]+c

    Author of lecture Basic Integration: Exercise 2

     Batool Akmal

    Batool Akmal

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