2011 Calculus AB free response #4a | AP Calculus AB | Khan Academy
TLDRThe video script discusses a continuous function f defined on the interval from -4 to 3, consisting of two quarter circles and a line segment. The function g(x) is introduced as 2x plus the definite integral of f(t) from 0 to x. The focus is on finding g(-3) and g'(x). Through careful analysis, it's determined that g(-3) equals -6 minus 9pi/4, and g'(x) is simply 2 plus the function f(x). The challenge lies in understanding the integration bounds and the geometric interpretation of the quarter circle's area.
Takeaways
- π The function g(x) is defined as g(x) = 2x + β« from 0 to x of f(t) dt, where f(x) consists of two quarter circles and one line segment.
- π’ The domain of the function f is from -4 to 3, as mentioned in the script.
- π To find g(-3), substitute x with -3 in the function g(x), which results in 2*(-3) + β« from 0 to -3 of f(t) dt.
- π€ The integral part of g(-3) represents the area under the graph of f(t) from 0 to -3, but with swapped bounds of integration, which affects the sign of the result.
- π Swapping the bounds of integration from 0 to -3 to -3 to 0 changes the sign of the integral, resulting in -β« from -3 to 0 of f(t) dt.
- π The area under one quarter circle (from -3 to 0) is calculated using geometry, with the radius being 3, leading to an area of 9Ο/4.
- π The derivative of g(x), denoted as g'(x), is found by differentiating each part of g(x). The derivative of 2x is 2, and by the fundamental theorem of calculus, the derivative of the integral is f(x).
- π Evaluating g'(-3) gives us 2 + f(-3), where f(-3) is determined by the function definition, which is 0 at x = -3.
- π― The final result for g(-3) is -6 - (9Ο/4), and for g'(-3) is 2.
- π‘ The process emphasizes the importance of understanding the bounds of integration and how they affect the sign of the integral.
- π The problem-solving approach in the script highlights the application of calculus concepts, such as derivatives and integrals, in finding values of composite functions.
Q & A
What is the domain of the continuous function f?
-The domain of the function f is the interval from -4 to 3, inclusive.
How is the graph of function f composed?
-The graph of function f consists of two quarter circles and one line segment, as described in the script.
What is the function g(x) defined as?
-The function g(x) is defined as g(x) = 2x plus the definite integral from 0 to x of f(t) dt.
How do you find the value of g(-3)?
-To find g(-3), you substitute -3 for x in the function g(x), which results in 2*(-3) and then evaluate the definite integral from 0 to -3 of f(t) dt, considering the swapped bounds of integration.
What is the geometric significance of the integral from 0 to -3 of f(t) dt?
-The integral from 0 to -3 of f(t) dt represents the area under the graph of f(t) from 0 to -3. Since f(t) is a quarter circle, the area can be calculated using geometry, which is 9Ο/4.
How do you calculate the area of the quarter circle in the integral?
-The area of a full circle with radius 3 is Ο*r^2, which is 9Ο. Since the quarter circle is 1/4 of the full circle, the area is 9Ο/4.
What is the derivative of g(x) with respect to x?
-The derivative of g(x), denoted as g'(x), is the derivative of the function 2x plus the definite integral from 0 to x of f(t) dt. By the fundamental theorem of calculus, g'(x) = 2 + f(x).
What is the value of g'(-3)?
-To find g'(-3), you evaluate the derivative of g(x) at x = -3. Since f(-3) is 0, g'(-3) equals 2 + f(-3), which is 2.
Why is it important to consider the bounds of integration when calculating the integral from 0 to -3 of f(t) dt?
-It is crucial because the order of the bounds affects the sign of the integral. When the lower bound is greater than the upper bound, the integral is negative. This understanding is essential for correctly evaluating the area under the curve.
What is the significance of the function f(x) at x = -3 in this context?
-The value of the function f(x) at x = -3 is important because it is used to evaluate g'(-3). In this case, f(-3) is 0, which simplifies the calculation of the derivative at that point.
How does the fundamental theorem of calculus apply to this problem?
-The fundamental theorem of calculus is used to find the derivative of the definite integral from 0 to x of f(t) dt. It states that this derivative is simply the function f(x) evaluated at x.
Outlines
π Mathematical Analysis of Function f and its Derived Function g
The paragraph introduces a continuous function f defined on the interval from -4 to 3, represented graphically by two quarter circles and a line segment. It then defines a new function g(x) as 2x plus the definite integral of f(t) from 0 to x. The primary task is to find g(-3), which involves substituting -3 into the function and calculating the integral from 0 to -3 of f(t), which is portrayed as an area under the graph of f. The integral is computed by considering the geometric area of a quarter circle with radius 3, resulting in an area of 9Ο/4. The paragraph also discusses finding the derivative g'(x), which is simply the derivative of the function 2x and the integral part, leading to g'(x) = 2 + f(x). The derivative at -3, g'(-3), is then evaluated, noting that f(-3) is 0, thus resulting in g'(-3) = 2. The paragraph emphasizes the importance of correctly interpreting the bounds of integration and the geometric interpretation of the integral.
Mindmap
Keywords
π‘continuous function
π‘interval
π‘quarter circles
π‘line segment
π‘definite integral
π‘area under the curve
π‘geometric calculation
π‘derivative
π‘fundamental theorem of calculus
π‘integration bounds
π‘function evaluation
Highlights
The function f is defined on the interval from negative 4 to 3 and consists of two quarter circles and one line segment.
The graph of f is visualized with one quarter circle on each side and a line segment in between, as depicted in the figure above.
The function g(x) is defined as 2x plus the definite integral from 0 to x of f(t) dt, which is an interesting composition of f(x).
To find g(-3), we substitute x with -3 in the definition of g(x), resulting in 2*(-3) plus the integral from 0 to -3 of f(t) dt.
The first part of calculating g(-3) is straightforward, yielding 2*(-3) which equals -6.
The integral part of g(-3) represents the area under f(t) from 0 to -3, which requires careful handling of the bounds of integration.
By swapping the bounds of integration from 0 to -3 to -3 to 0, the sign of the integral changes, leading to the expression - integral from -3 to 0 of f(t) dt.
The area under the quarter circle can be calculated using geometry, with the radius being 3 and the entire circle having an area of pi r squared.
The area under the quarter circle is 1/4 of the entire circle's area, which is 9 pi / 4.
The final result for g(-3) is -6 - 9 pi / 4, combining the numerical and integral parts.
For Part a, we also need to find g'(x), which is the derivative of g(x), and starts with the derivative of 2x, which is 2.
The derivative of the definite integral part of g(x) is f(x) itself, by the fundamental theorem of calculus.
Evaluating g'(x) at -3 gives us 2 plus f(-3), where f(-3) is found to be 0 from the function definition.
The value of g'(-3) is thus 2, as 2 plus f(-3) which is 0 equals 2.
The key takeaway from Part a is understanding the bounds of integration and how they affect the calculated area.
The integral from 0 to -3 is not simply the area under the curve in that interval but the negative of that area due to the swapped bounds.
The problem-solving process in Part a involves both algebraic manipulation and geometric understanding to arrive at the correct solution.
Transcripts
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