Integrate (6x^2-cos(2x))dx??
Using the "Just look at it Theorem"
Integral (6x^2 - cos(2x) dx
= 2x^3 - (1/2)sin(2x) + constant
you can easily check by mentally finding the derivative of the above answer
That's my favorite theorem! Too bad it's not taught more.
In all fairness, when you're brand new at this stuff, things often seem more confusing than they are.
The main problem I encounter is that students seem to forget what the exercise set covers. If you've just been studying integration of powers and trig functions, it's not likely that the problems will branch out into strange territory.
Same thing with word problems. Read the story -- it will almost always produce an equation just like those you just worked in the previous section of exercises!!!
To integrate the expression (6x^2 - cos(2x)), we'll break it down into two separate integrals. One for the term 6x^2, and another for the term cos(2x).
Let's start with the integral of 6x^2. To integrate x raised to a power, we increase the power by 1 and divide by the new power. So, the integral of x^2 would be (1/3)x^3.
Now, let's move on to the integral of cos(2x). The integral of cosine functions can be found using the sine function. The integral of cos(x) is sin(x), and since we have cos(2x), we need to divide by the coefficient of x, which in this case is 2. Thus, the integral of cos(2x) is (1/2)sin(2x).
Now we can write the final result by adding the integrals of the two terms:
∫ (6x^2 - cos(2x)) dx = ∫ 6x^2 dx - ∫ cos(2x) dx
The integral of 6x^2 is (1/3)x^3, and the integral of cos(2x) is (1/2)sin(2x).
So, our final result is:
∫ (6x^2 - cos(2x)) dx = (1/3)x^3 - (1/2)sin(2x) + C
where C represents the constant of integration.