Question

1. given the graph of $f(x)$ below, what is $int_{3}^{11} f(x) \\, dx$? note that the arc from $x = 7$ to $x = 11$ is a quarter - circle centered at $(7,0)$.

graph of the function $f(x)$ with a grid, showing a line from $(0,2)$ to $(4, - 2)$, then a line to $(7,4)$ and a quarter - circle from $x = 7$ to $x = 11$ centered at $(7,0)$
(a) $\frac{1}{2}+16\pi$
(b) $\frac{\pi + 3}{2}$
(c) $4\pi+7$
(d) $4\pi + 3$
(e) $\frac{8\pi+3}{2}$

1. suppose that $f(x)=\sin(\pi)+x$ and $f(\pi)=\pi^{2}+1$. what is $f\left(\frac{\pi}{2}\

ight)$?
(a) $-\frac{3\pi^{2}}{8}$
(b) $\frac{\pi^{2}}{5}$
(c) $\frac{3\pi^{2}}{8}$
(d) $\frac{\pi^{2}}{8}+2$
(e) $\frac{\pi^{2}}{6}+2$

Explanation:

Question 3

Step1: Split the integral into parts

The integral \(\int_{3}^{11} f(x) \, dx\) can be split into three parts: from \(x = 3\) to \(x = 4\), \(x = 4\) to \(x = 7\), and \(x = 7\) to \(x = 11\).

Step2: Calculate the area from \(x = 3\) to \(x = 4\)

This is a triangle with base \(1\) (from \(3\) to \(4\)) and height \(2 - (-2) = 4\)? Wait, no. Wait, from \(x = 3\) to \(x = 4\), the graph goes from \(y = 0\) (at \(x = 3\))? Wait, no, looking at the graph: at \(x = 3\), the point is on the line from \(x = 0\) (wait, no, the graph starts at \(x = 0\) with \(y = 2\), then goes down to \(x = 4\) with \(y = -2\). Wait, the first segment is from \(x = 0\) to \(x = 4\), but our integral starts at \(x = 3\) to \(x = 4\). So from \(x = 3\) to \(x = 4\), the function is a line from \((3, 0)\)? Wait, no, maybe I misread. Wait, the graph: at \(x = 0\), \(y = 2\); at \(x = 4\), \(y = -2\). Then from \(x = 4\) to \(x = 7\), it's a line going up to \((7, 4)\)? Wait, no, the graph from \(x = 4\) to \(x = 7\) is a line from \((4, -2)\) to \((7, 4)\)? Wait, no, at \(x = 7\), the \(y\)-value is \(4\)? Wait, the arc from \(x = 7\) to \(x = 11\) is a quarter-circle centered at \((7, 0)\). So the center is \((7, 0)\), so the radius is the distance from \((7, 0)\) to the top of the arc, which is \(4\) (since at \(x = 7\), \(y = 4\)? Wait, no, if the center is \((7, 0)\), then the quarter-circle is in the upper half (since \(y\) is positive there). So the radius \(r = 4\) (from \(y = 0\) to \(y = 4\) at \(x = 7\)).

Wait, let's re-examine:

1. From \(x = 3\) to \(x = 4\): Wait, maybe the first part is from \(x = 3\) to \(x = 4\): the graph is a triangle? Wait, no, from \(x = 0\) to \(x = 4\), the line goes from \((0, 2)\) to \((4, -2)\). So the equation of this line is \(y = -x + 2\) (since at \(x = 0\), \(y = 2\); slope is \(\frac{-2 - 2}{4 - 0} = -1\), so \(y = -x + 2\)). At \(x = 3\), \(y = -3 + 2 = -1\)? Wait, maybe I'm overcomplicating. Alternatively, the area from \(x = 3\) to \(x = 4\): the region is a triangle with base \(1\) (from \(3\) to \(4\)) and height \(|y|\) at \(x = 3\) and \(x = 4\). Wait, maybe better to split the integral into:

• From \(x = 3\) to \(x = 4\): area of a triangle (or trapezoid)
• From \(x = 4\) to \(x = 7\): area of a trapezoid or triangle
• From \(x = 7\) to \(x = 11\): area of a quarter-circle

Wait, let's correct:

First, from \(x = 3\) to \(x = 4\): the function is part of the line from \(x = 0\) to \(x = 4\) (since \(3\) to \(4\) is within \(0\) to \(4\)). The line from \((0, 2)\) to \((4, -2)\) has equation \(y = -x + 2\). At \(x = 3\), \(y = -1\); at \(x = 4\), \(y = -2\). So the area from \(x = 3\) to \(x = 4\) is the area under the curve (but since it's below the x-axis? Wait, no, the integral is the net area, but actually, we need to calculate the area (signed or not? Wait, the integral of \(f(x) dx\) is the net area, but since we are calculating the area (as the graph is above and below, but we need to consider the sign). Wait, maybe better to calculate the area as the sum of the areas of the regions, considering their signs (positive above x-axis, negative below).

Wait, from \(x = 3\) to \(x = 4\): the region is a triangle with base \(1\) (from \(3\) to \(4\)) and height \(| -2 - (-1) | = 1\)? No, this is getting confusing. Maybe another approach:

Wait, the integral from \(3\) to \(11\) can be split into:

1. From \(3\) to \(4\): a triangle (or part of a triangle)
2. From \(4\) to \(7\): a trapezoid
3. From \(7\) to \(11\): a quarter-circle (but since it's a quarter-circle centered at \((7, 0)\), the radius…

Answer:

Step1: Split the integral into parts

The integral \(\int_{3}^{11} f(x) \, dx\) can be split into three parts: from \(x = 3\) to \(x = 4\), \(x = 4\) to \(x = 7\), and \(x = 7\) to \(x = 11\).

Step2: Calculate the area from \(x = 3\) to \(x = 4\)

This is a triangle with base \(1\) (from \(3\) to \(4\)) and height \(2 - (-2) = 4\)? Wait, no. Wait, from \(x = 3\) to \(x = 4\), the graph goes from \(y = 0\) (at \(x = 3\))? Wait, no, looking at the graph: at \(x = 3\), the point is on the line from \(x = 0\) (wait, no, the graph starts at \(x = 0\) with \(y = 2\), then goes down to \(x = 4\) with \(y = -2\). Wait, the first segment is from \(x = 0\) to \(x = 4\), but our integral starts at \(x = 3\) to \(x = 4\). So from \(x = 3\) to \(x = 4\), the function is a line from \((3, 0)\)? Wait, no, maybe I misread. Wait, the graph: at \(x = 0\), \(y = 2\); at \(x = 4\), \(y = -2\). Then from \(x = 4\) to \(x = 7\), it's a line going up to \((7, 4)\)? Wait, no, the graph from \(x = 4\) to \(x = 7\) is a line from \((4, -2)\) to \((7, 4)\)? Wait, no, at \(x = 7\), the \(y\)-value is \(4\)? Wait, the arc from \(x = 7\) to \(x = 11\) is a quarter-circle centered at \((7, 0)\). So the center is \((7, 0)\), so the radius is the distance from \((7, 0)\) to the top of the arc, which is \(4\) (since at \(x = 7\), \(y = 4\)? Wait, no, if the center is \((7, 0)\), then the quarter-circle is in the upper half (since \(y\) is positive there). So the radius \(r = 4\) (from \(y = 0\) to \(y = 4\) at \(x = 7\)).

Wait, let's re-examine:

1. From \(x = 3\) to \(x = 4\): Wait, maybe the first part is from \(x = 3\) to \(x = 4\): the graph is a triangle? Wait, no, from \(x = 0\) to \(x = 4\), the line goes from \((0, 2)\) to \((4, -2)\). So the equation of this line is \(y = -x + 2\) (since at \(x = 0\), \(y = 2\); slope is \(\frac{-2 - 2}{4 - 0} = -1\), so \(y = -x + 2\)). At \(x = 3\), \(y = -3 + 2 = -1\)? Wait, maybe I'm overcomplicating. Alternatively, the area from \(x = 3\) to \(x = 4\): the region is a triangle with base \(1\) (from \(3\) to \(4\)) and height \(|y|\) at \(x = 3\) and \(x = 4\). Wait, maybe better to split the integral into:

• From \(x = 3\) to \(x = 4\): area of a triangle (or trapezoid)
• From \(x = 4\) to \(x = 7\): area of a trapezoid or triangle
• From \(x = 7\) to \(x = 11\): area of a quarter-circle

Wait, let's correct:

First, from \(x = 3\) to \(x = 4\): the function is part of the line from \(x = 0\) to \(x = 4\) (since \(3\) to \(4\) is within \(0\) to \(4\)). The line from \((0, 2)\) to \((4, -2)\) has equation \(y = -x + 2\). At \(x = 3\), \(y = -1\); at \(x = 4\), \(y = -2\). So the area from \(x = 3\) to \(x = 4\) is the area under the curve (but since it's below the x-axis? Wait, no, the integral is the net area, but actually, we need to calculate the area (signed or not? Wait, the integral of \(f(x) dx\) is the net area, but since we are calculating the area (as the graph is above and below, but we need to consider the sign). Wait, maybe better to calculate the area as the sum of the areas of the regions, considering their signs (positive above x-axis, negative below).

Wait, from \(x = 3\) to \(x = 4\): the region is a triangle with base \(1\) (from \(3\) to \(4\)) and height \(| -2 - (-1) | = 1\)? No, this is getting confusing. Maybe another approach:

Wait, the integral from \(3\) to \(11\) can be split into:

1. From \(3\) to \(4\): a triangle (or part of a triangle)
2. From \(4\) to \(7\): a trapezoid
3. From \(7\) to \(11\): a quarter-circle (but since it's a quarter-circle centered at \((7, 0)\), the radius is \(4\) (since from \(x = 7\), the top of the arc is at \(y = 4\), so radius \(r = 4\)). A quarter-circle has area \(\frac{1}{4} \pi r^2 = \frac{1}{4} \pi (4)^2 = 4\pi\).

Now, from \(4\) to \(7\): the graph is a line from \((4, -2)\) to \((7, 4)\). The area under this line (from \(x = 4\) to \(x = 7\)) is a trapezoid with bases \(|-2| = 2\) and \(4\), and height \(7 - 4 = 3\). Wait, no, the area of a trapezoid is \(\frac{1}{2} (b_1 + b_2) h\). Here, \(b_1 = 2\) (the length at \(x = 4\), but since it's below the x-axis, it's negative, and \(b_2 = 4\) (above the x-axis). Wait, actually, the net area would be the area of the trapezoid: \(\frac{1}{2} (2 + 4) \times 3 = 9\)? Wait, no, because from \(x = 4\) to \(x = 7\), the function goes from \(y = -2\) to \(y = 4\), so the area between the curve and the x-axis is a trapezoid with bases \(2\) (the vertical distance from x-axis to \(y = -2\)) and \(4\) (vertical distance from x-axis to \(y = 4\)), and horizontal length \(3\) (from \(4\) to \(7\)). But since part is below and part is above, we can calculate the area as the area of the triangle below the x-axis and the triangle above. Wait, from \(x = 4\) to \(x = 5\) (where \(y = 0\)): the line from \((4, -2)\) to \((7, 4)\) has slope \(\frac{4 - (-2)}{7 - 4} = 2\). So equation: \(y - (-2) = 2(x - 4)\) => \(y = 2x - 10\). Setting \(y = 0\), \(2x - 10 = 0\) => \(x = 5\). So from \(x = 4\) to \(x = 5\), the area is a triangle with base \(1\) (from \(4\) to \(5\)) and height \(2\) (from \(y = -2\) to \(y = 0\)), area \(\frac{1}{2} \times 1 \times 2 = 1\) (but below x-axis, so negative? Wait, no, the integral is the net area, so areas below x-axis are negative, above are positive. Wait, no, actually, the integral \(\int_{a}^{b} f(x) dx\) is the area above the x-axis minus the area below the x-axis. So from \(x = 4\) to \(x = 5\), the function is negative (below x-axis), so area is \(-\frac{1}{2} \times 1 \times 2 = -1\). From \(x = 5\) to \(x = 7\), the function is positive (above x-axis), with base \(2\) (from \(5\) to \(7\)) and height \(4\) (at \(x = 7\)), so area is \(\frac{1}{2} \times 2 \times 4 = 4\). So total from \(x = 4\) to \(x = 7\) is \(-1 + 4 = 3\)? Wait, no, that's not right. Wait, the trapezoid area formula: \(\frac{1}{2} (b_1 + b_2) h\), where \(b_1\) and \(b_2\) are the lengths of the two parallel sides (the y-values at the endpoints). So \(b_1 = -2\) (at \(x = 4\)) and \(b_2 = 4\) (at \(x = 7\)), and \(h = 3\) (the horizontal distance). So area is \(\frac{1}{2} (-2 + 4) \times 3 = \frac{1}{2} \times 2 \times 3 = 3\). That's correct, because the net area is the average of the two y-values times the width.

Now, from \(x = 3\) to \(x = 4\): the function at \(x = 3\) is \(y = -1\) (from \(y = -x + 2\), \(x = 3\) gives \(y = -1\)) and at \(x = 4\) is \(y = -2\). So the area is \(\frac{1}{2} (-1 + (-2)) \times 1 = \frac{1}{2} (-3) \times 1 = -\frac{3}{2}\)? Wait, no, the width is \(1\) (from \(3\) to \(4\)), and the average of the two y-values is \(\frac{-1 + (-2)}{2} = -\frac{3}{2}\), so area is \(-\frac{3}{2}\).

Now, from \(x = 7\) to \(x = 11\): the quarter-circle centered at \((7, 0)\) with radius \(4\) (since the top of the arc is at \(y = 4\), so radius \(r = 4\)). The area of a quarter-circle is \(\frac{1}{4} \pi r^2 = \frac{1}{4} \pi (4)^2 = 4\pi\). Since it's above the x-axis, this area is positive.

Now, summing up the three parts:

• From \(3\) to \(4\): \(-\frac{3}{2}\)
• From \(4\) to \(7\): \(3\)
• From \(7\) to \(11\): \(4\pi\)

Wait, but that can't be right. Wait, maybe I made a mistake in the first part. Wait, the graph from \(x = 0\) to \(x = 4\) is a line from \((0, 2)\) to \((4, -2)\). So at \(x = 3\), the y-value is \(y = -3 + 2 = -1\) (correct). From \(x = 3\) to \(x = 4\), the area under the curve (integral) is the integral of \(y = -x + 2\) from \(3\) to \(4\). The integral of \(-x + 2\) is \(-\frac{1}{2}x^2 + 2x\). Evaluated from \(3\) to \(4\):

At \(x = 4\): \(-\frac{1}{2}(16) + 8 = -8 + 8 = 0\)

At \(x = 3\): \(-\frac{1}{2}(9) + 6 = -4.5 + 6 = 1.5\)

So the integral from \(3\) to \(4\) is \(0 - 1.5 = -1.5 = -\frac{3}{2}\) (correct).

From \(x = 4\) to \(x = 7\), the function is a line from \((4, -2)\) to \((7, 4)\). The equation is \(y = 2x - 10\) (as before). The integral of \(2x - 10\) from \(4\) to \(7\) is \(x^2 - 10x\) evaluated from \(4\) to \(7\).

At \(x = 7\): \(49 - 70 = -21\)

At \(x = 4\): \(16 - 40 = -24\)

So the integral is \(-21 - (-24) = 3\) (correct).

From \(x = 7\) to \(x = 11\), the function is the quarter-circle centered at \((7, 0)\) with radius \(4\). The equation of the circle is \((x - 7)^2 + y^2 = 16\), so \(y = \sqrt{16 - (x - 7)^2}\) (since it's the upper half). The integral of \(y\) from \(7\) to \(11\) is the area of the quarter-circle, which is \(\frac{1}{4} \pi (4)^2 = 4\pi\) (correct).

Now, summing all three parts: \(-\frac{3}{2} + 3 + 4\pi = \frac{3}{2} + 4\pi\). Wait, but the options are:

(a) \(\frac{1}{2} + 16\pi\) – no

(b) \(\frac{\pi + 3}{2}\) – no

(c) \(4\pi + 7\) – no

(d) \(4\pi + 3\) – no

(e) \(\frac{8\pi + 3}{2}\) – wait, maybe I split the integral wrong. Wait, maybe the first part is from \(x = 3\) to \(x = 4\) is not part of the first line. Wait, maybe the graph from \(x = 0\) to \(x = 4\) is a triangle with base \(4\) and height \(4\) (from \(y = 2\) to \(y = -2\)), but our integral starts at \(x = 3\) to \(x = 4\), which[SSE Completed, Client Connection Error][LLM SSE On Failure]