Unit 8 practice test: Applications of Integration

On $[0, 2]$ the graph of $f$ is on top, so each vertical rectangle has height $f(x) - g(x)$, and the curves meet at $x = 0$ and $x = 2$.

• Correct: $\int_0^2 \left[(3 + 2x - x^2) - (x^2 - 2x + 3)\right]\,dx$: Top minus bottom, from the left crossing to the right one.
• $\int_0^2 \left[(x^2 - 2x + 3) - (3 + 2x - x^2)\right]\,dx$: Bottom minus top: every rectangle gets a negative height, and the integral comes out as $-\frac{8}{3}$.
• $\int_0^2 (3 + 2x - x^2)\,dx$: This is the area under $f$ all the way down to the $x$-axis, which includes the space under $g$.
• $\int_2^4 \left[(3 + 2x - x^2) - (x^2 - 2x + 3)\right]\,dx$: The limits are the $y$-values 2 and 4. With $dx$ the limits are $x$-values: the crossings at $x = 0$ and $x = 2$.

The region is closed where the curves meet. $3 + 2x - x^2$$= x^2 - 2x + 3$ gives $2x^2 - 4x = 0$, so $x = 0$ or $x = 2$.

• Correct: Solve $f(x) = g(x)$, which gives $x = 0$ and $x = 2$: The curves meet where $f(x) = g(x)$; those $x$-values close the region.
• Solve $f(x) = 0$, which gives where $f$ meets the $x$-axis: Where $f$ meets the $x$-axis matters for the area under $f$, not for the region between two curves.
• Use the $y$-values where the curves meet, 3 and 3: An integral in $dx$ runs between $x$-values. The points meet at $y = 3$, but the limits are their $x$-coordinates.
• Use the lowest and highest points of $R$, $y = 2$ and $y = 4$: Heights of the region are not limits. The limits come from where the curves cross, at $x$-values.

The height is $h(x) = 4x - 2x^2$, with $h'(x) = 4 - 4x = 0$ at $x = 1$, and $h'$ changes from positive to negative there. $h(1) = 4 - 2 = 2$.

• Correct: At $x = 1$, where $f(x) - g(x) = 2$: $h(1) = 2$ is the largest value of $h(x) = 4x - 2x^2$ on $[0, 2]$.
• At $x = 1$, where $f(x) = 4$: 4 is the height of the top curve above the $x$-axis. The region's height is top minus bottom: $4 - 2 = 2$.
• At $x = 2$, where the curves meet: Where the curves meet, the height of $R$ is 0, its smallest.
• At $x = 0$, where $f(x) - g(x) = 3$: At $x = 0$ both curves equal 3, so the height there is 0, not 3.

Slice horizontally: each strip runs from the parabola on the left to the line on the right, so its length is right minus left, and $y$ runs from $-2$ to $4$.

• Correct: $\int_{-2}^{4} \left[(y + 4) - \frac{y^2}{2}\right]\,dy$: Right minus left, in $dy$, between the $y$-values where the curves meet.
• $\int_{-2}^{4} \left[\frac{y^2}{2} - (y + 4)\right]\,dy$: Left minus right gives a negative number. The line $x = y + 4$ is the right edge.
• $\int_{0}^{8} \left[\sqrt{2x} - (x - 4)\right]\,dx$: With vertical slices the bottom edge changes at $x = 2$: from $x = 0$ to $2$ the bottom is $y = -\sqrt{2x}$, not $y = x - 4$. One integral in $x$ is not enough.
• $\int_{2}^{8} \left[(y + 4) - \frac{y^2}{2}\right]\,dy$: 2 and 8 are the $x$-coordinates of the crossings. An integral in $dy$ needs the $y$-coordinates, $-2$ and $4$.

$\frac{y^2}{2} = y + 4$ gives $y^2 - 2y - 8 = 0$, so $(y - 4)(y + 2) = 0$: $y = -2$ and $y = 4$.

• Correct: $y = -2$ and $y = 4$, where $\frac{y^2}{2} = y + 4$: In $dy$, the limits are the lowest and highest $y$ of the region.
• $x = 2$ and $x = 8$, the $x$-values where the curves meet: These are $x$-values. With $dy$ the limits must be $y$-values.
• $y = 0$ and $y = 4$, from the $x$-axis up to the top: The region goes below the $x$-axis, down to $y = -2$.
• $x = 0$ and $x = 8$, from the left edge to the right: These are $x$-values, and the integral is in $dy$.

Every horizontal strip runs from the same left curve to the same right curve. Vertical strips end on $y = -\sqrt{2x}$ for $0 \le x \le 2$ and on $y = x - 4$ for $2 \le x \le 8$, so they need two integrals.

• Correct: Vertical strips change bottom curves at $x = 2$: One formula for each edge in $y$; two pieces in $x$.
• Horizontal strips give a larger and more accurate area: Both ways give the same area, 18. The choice only changes how much work it takes.
• The region lies to the right of the $y$-axis, where $x > 0$: Where the region sits does not decide the method. What matters is whether each edge is one formula.
• Vertical strips would make the area of $R$ negative: Done correctly, vertical strips give $+18$ too, with two integrals.