Understanding Integrals: Analyzing Graphs and Practice Problems

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SUMMARY

This discussion focuses on the analysis of integrals and their graphical representations, specifically addressing the computation of definite integrals and the interpretation of derivative behavior. The user correctly computes the integral $$g(0)=\int_1^0 f(t)\,dt$$ and identifies the tangent line at a point on the graph. Additionally, the discussion highlights the significance of relative maxima in the context of the function's derivative, with specific reference to $$g'(-1)=0$$ indicating a maximum point. The conversation emphasizes the importance of understanding both numeric answers and the underlying graphical interpretations.

PREREQUISITES
  • Understanding of definite integrals and their properties
  • Familiarity with the concept of derivatives and their graphical implications
  • Knowledge of tangent lines and their equations
  • Ability to interpret graphs of functions and their derivatives
NEXT STEPS
  • Study the Fundamental Theorem of Calculus for deeper insights into integrals
  • Learn about the graphical interpretation of derivatives using software like Desmos
  • Explore relative extrema and their significance in calculus
  • Practice solving integrals and analyzing their graphs using tools like Wolfram Alpha
USEFUL FOR

Students and educators in calculus, mathematicians interested in integral analysis, and anyone seeking to improve their understanding of graphical interpretations of functions and their derivatives.

karush
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View attachment 2234

just see if I did this right
new stuff for me
the graph and typing is mine
thanks much ahead
 
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a) You do have the correct numeric answer, but I don't understand why you have computed those other integrals. I would simply write:

$$g(0)=\int_1^0 f(t)\,dt=-\int_0^1 f(t)\,dt=-(-2)=2$$

b) Again, you have the correct numeric value, but I would write instead:

The tangent line is:

$$y-g(3)=g'(3)(x-3)$$

$$y+3=0$$

$$y=-3$$

c)

A) $$g'(-1)=0$$ To the left of $x=-1$ we see that $g'(x)>0$ and to the right of $x=-1$ we see that $g'(x)<0$, hence $g(x)$ has a relative maximum there.

B) Use similar reasoning as part A).

C) $g'(-2)\ne0$...

D) Correct, but why?

E) Correct, but why?
 
See what you mean..
I still have a hard time looking at these graphs and see what is going on
More to come..
 

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