Let g(x)=∫0^x f(t) d t, where f is the function whose graph is shown. (a) Evaluate g(0), g

Let g(x)=∫0^x f(t) d t, where f is the function whose graph...

Key Concepts

Critical Points and Maximum Values

A critical point in an accumulation function occurs where its derivative (which is f(x)) is zero or changes sign. These points are important for identifying local extrema, such as maximum and minimum values, of the accumulation function g(x). Recognizing where f changes from positive to negative helps in pinpointing the location of g's maximum value.

Monotonicity of a Function

The concept of monotonicity refers to the intervals where a function is consistently increasing or decreasing. In the case of an accumulation function, g(x) increases when f(x) is positive and decreases when f(x) is negative. This relationship helps in determining the intervals of growth or decline in the accumulated area.

Accumulation Functions

An accumulation function, like g(x) = ??? f(t) dt, measures the net signed area under the curve of f from a fixed start point to x. It captures how quantities accumulate over an interval and is useful for translating local behavior (values of f) into the cumulative behavior described by g.

Fundamental Theorem of Calculus

This theorem connects differentiation and integration, stating that if a function is integrated and the resulting accumulation function is differentiated, the original function is recovered. It underpins the idea that the derivative of the integral function g(x) is the original function f(x), and it is essential for understanding how changes in f affect g.

Transcript

00:01 Function g of x equal to integral f of t d t 0 to x where f is the function whose graph is shown you can see that here so we have to evaluate g of 0 g of 1, g of 2, g of 3 and g of 6.

00:29 So first of all we have to see that in this graph as g of x is defined it is the area between graph of f and d axis x equal to 0 to x equal to t because of the way f is shaped we can get exit value because some of the area are half rectangle so you can see that here x equal to zero so g of equal to integral f of t d t d t 0 to 0 so here we know that when limit 0 equal to 0 then we get here is 0 because the integral with equal limits so we get here 0 when x equal to 1 then g of 1 equal to integral f of t d t 0 to 1 so here you can see that it is a rectangle and we know that bass is 1 and height is 2 so we get here after that we have to find 4 x equal to 2 so here g of 2 equal to integral 0 to 2 f of t d t now you can see that here integral 0 to 1 f of t d t so we know that the value of f of t d t so we know that the value of f of t d t integral f of t t 0 to 1 which is 2 so we put here 2 plus 1 to 2 you can see that we have rectangle and here is we have triangle so we know that for triangle base is 1 and hide is 2 so we put here plus 4 triangle we know that 1 1 upon 2 base is 1 and height is 2 so now we simplify that then we get g of 2 equal to 5 after that when x equal to 3 so g of 3 equal to integral f of t d t 0 to 3 so here we know that integral f of t d t 0 to 2 plus integral f of t d t 2 2 2 so you can see that you know that the value of integral f of t d t 0 to 2 which is 5 so we put here 5 plus now you can see that we have a triangle 2 to 3 f of t d t so we we get one upon two base is one and height is four so now we simplify that and we get seven so g of three equal to seven now you can see that we have to find g of six so x equal to six so we have to find g of six equal to integral 0 to 6 f of t d t equal to integral 0 to 3 f of t d t plus integral 3 to 4 f of t d t plus integral integral f of t d t 4 to 5 plus integral f of t d t 5 to 6 so you can see that here integral f of t d t 0 to 3 which is 7 plus you can see that integral f of t d t 3 which is 7 plus you can see that integral f of t d t 3 to 4 which is here and you can can see that it is a triangle and bass is one and height is negative one so we put here upon two that is a formula of triangle plus integral f of t d t 4 to so which is here and here is rectangle and triangle one time one plus four triangle 1 upon 2 time base is 1 and height is negative 1 so we put here and here is you can see that height is negative 1 so we put here negative 1 plus for integral up of t d t 5 to 6 so you can see that rectangle here is base is 1 and height is negative 2 so put here now we simplify this and we get g of 6 equal to 7 negative 4 equal to 3 so g of 6 equal to c so you can see that here g of 0 equal to 0 g of 1 equal to 2 g of 2 equal to 2 g of 2 equal to 5 g of 3 equal to 7 and g of 6 equal to 3.

08:13 So it is our final answer for part a.

08:19 Now we solve part b...

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