How do you integrate equations where f goes from some value to x?

Question

anonymous · Answer

@sanhitha why don't you post specific problem, instead of asking vague post?

anonymous · Answer

okay fine ...sorry|dw:1335108532304:dw|

anonymous · Answer

anybody??

ash2326 · Answer

Is  this your question?
$$f(x) =\int_{1} ^{x} (x^2-2x ) dx$$

anonymous · Answer

yupp

ash2326 · Answer

Ok , whenever you have this type of function
As you know that integral is inverse of derivative so, here we won't need to differentiate explicitly
$$f(x)=\int_{1} ^x (x^2-2x) dx$$
$$f'(x)=\frac{d}{dx}(\int_1 ^{x} (x^2-2x) dx$$

ash2326 · Answer

To find f'(x) plug in upper limit and then multiply this by the derivative of lower limit, subtract this by the lower limit plugged in and product by its derivative 
$$f'(x)= (x^2-2x) \frac{d}{dx} (x)-(1^2-2)\frac{d}{dx} (1)$$
So we get
$$f'(x)= (x^2-2x) \times 1-0$$

ash2326 · Answer

Sorry
To find f'(x) plug in upper limit and then multiply this by the derivative of upper limit, subtract this by the lower limit plugged in and product by its derivative