4.1.306 AP Calculus Exam Area under Curve

In summary: That's not what was asked in the question. The question was asking for the area of the region, which is simply the integral of the function $e^{x/2}$ from $x=0$ to $x=2$. So in summary, the area of the region bounded by $y=e^{x/2}$ and $x=2$ in the first quadrant is 2e-2.
  • #1
karush
Gold Member
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$\textsf{What is the area of the region in the first quadrant bounded by the graph of}$
$$y=e^{x/2} \textit{ and the line } x=2$$
a. 2e-2 b. 2e c. $\dfrac{e}{2}-1$ d. $\dfrac{e-1}{2}$ e. e-1Integrate
$\displaystyle \int e^{x/2}=2e^{x/2}$
take the limits
$2e^{x/2}\Biggr|_0^2=2e-2$which is a.any suggestions ?
 
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  • #2
karush said:
$\textsf{What is the area of the region in the first quadrant bounded by the graph of}$
$$y=e^{x/2} \textit{ and the line } x=2$$
a. 2e-2 b. 2e c. $\dfrac{e}{2}-1$ d. $\dfrac{e-1}{2}$ e. e-1Integrate
$\displaystyle \int e^{x/2}=2e^{x/2}$
take the limits
$2e^{x/2}\Biggr|_0^2=2e-2$which is a.any suggestions ?

Put a +C at the end of your indefinite integral, and then you are correct.
 
  • #3
karush said:
Integrate
\(\displaystyle \int e^{x/2}\) dx \(\displaystyle =2e^{x/2}\) + C
Tsk tsk.

-Dan
 
  • #4
\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[axis lines=middle, grid=both]
\addplot[
draw = red,
domain=-.5:3,
range=-2:8,
samples=50
] {exp(e^(x/2)};
\draw[solid] (2,0)- -(2,6);
\end{axis}
\end{tikzpicture}ok I tried to tikx $e^{x/2}$ and $x=2$ but kinda ! don't see any vertical line
also be nice for a light shade in the region
 
  • #5
karush said:
\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[axis lines=middle, grid=both]
\addplot[
draw = red,
domain=-.5:3,
range=-2:8,
samples=50
] {exp(e^(x/2)};
\draw[solid] (2,0)- -(2,6);
\end{axis}
\end{tikzpicture}ok I tried to tikx $e^{x/2}$ and $x=2$ but kinda ! don't see any vertical line
also be nice for a light shade in the region

Wait, what? Are you trying to graph the area?
 

Related to 4.1.306 AP Calculus Exam Area under Curve

1. What is the significance of the "4.1.306" in the title of the AP Calculus Exam?

The "4.1.306" refers to the section, subsection, and question number in the AP Calculus curriculum. In this case, it is the fourth section, first subsection, and 306th question in the curriculum related to area under a curve.

2. What does the term "area under curve" mean in the context of calculus?

The area under a curve is the total area between a curve and the x-axis on a graph. In calculus, it is used to calculate the exact value of the area under a curve by taking the limit of smaller and smaller rectangles that approximate the area.

3. How is the concept of area under a curve related to integration?

Integration is the mathematical process used to find the area under a curve. It involves finding the antiderivative of a function and evaluating it between two points on the x-axis. This value represents the area under the curve between those two points.

4. What is the formula for calculating the area under a curve?

The formula for calculating the area under a curve is given by the definite integral: ∫ab f(x) dx, where f(x) is the function representing the curve, and a and b are the lower and upper limits of integration, respectively.

5. How is the concept of area under a curve applied in real-world situations?

The concept of area under a curve is applied in many real-world situations, such as calculating the distance traveled by an object given its velocity over time, finding the total revenue of a business given its demand function, or determining the amount of medication in a patient's bloodstream over time. It is a fundamental concept in physics, economics, and many other fields.

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