Calculus BC ยท Unit 8
Subtract top minus bottom, or right minus left. Split the integral where the curves trade places.
Top minus bottom
For curves given as functions of \(x\),
\(\int_a^b \left(y_{\text{top}}-y_{\text{bottom}}\right)\,dx\)
Right minus left
For curves given as functions of \(y\),
\(\int_c^d \left(x_{\text{right}}-x_{\text{left}}\right)\,dy\)
Find the intersections before you integrate
The limits are where the curves meet, unless the problem gives the interval.
More than two intersections means more than one piece
Subtract in the order that stays nonnegative on each piece. A single integral of \(f-g\) can cancel area.
The limits are small, and one curve is clearly above the other.
On \([0,1]\), \(x\ge x^2\).
\(\int_0^1 (x-x^2)\,dx\)
\(\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1\)
\(\dfrac{1}{2}-\dfrac{1}{3}\)
\(\dfrac{1}{6}\)
Answer: \(\dfrac{1}{6}\)
\(\int_0^5 (4-1)\,dx\)
\(3\cdot 5\)
\(15\)
Answer: \(15\)
\(\int_0^2 6x\,dx\)
\(\left[3x^2\right]_0^2\)
\(12\)
Answer: \(12\)
\(\int_0^4 x^{1/2}\,dx\)
\(\left[\dfrac{2}{3}x^{3/2}\right]_0^4\)
\(\dfrac{2}{3}\cdot 8\)
\(\dfrac{16}{3}\)
Answer: \(\dfrac{16}{3}\)
At \(x=1\), \(2x=2\) and \(x^2=1\).
The line is on top.
Answer: \(y=2x\)
\(\int_0^2 (2x-x^2)\,dx\)
\(\left[x^2-\dfrac{x^3}{3}\right]_0^2\)
\(4-\dfrac{8}{3}\)
\(\dfrac{4}{3}\)
Answer: \(\dfrac{4}{3}\)
Use dy, or a region that is symmetric.
Intersections: \(4-x^2=0\), so \(x=-2\) and \(x=2\).
\(\int_{-2}^{2}(4-x^2)\,dx\)
The integrand is even, so use \(2\int_0^2\).
\(2\left[4x-\dfrac{x^3}{3}\right]_0^2\)
\(2\left(8-\dfrac{8}{3}\right)\)
\(2\cdot \dfrac{16}{3}\)
\(\dfrac{32}{3}\)
Answer: \(\dfrac{32}{3}\)
Intersections: \(y^2=y+2\), so \(y^2-y-2=0\).
\((y-2)(y+1)=0\)
\(y=-1\) and \(y=2\).
The line is to the right of the parabola on that interval.
\(\int_{-1}^{2}(y+2-y^2)\,dy\)
\(\left[\dfrac{y^2}{2}+2y-\dfrac{y^3}{3}\right]_{-1}^{2}\)
At 2: \(\dfrac{10}{3}\). At \(-1\): \(-\dfrac{7}{6}\).
\(\dfrac{10}{3}-\left(-\dfrac{7}{6}\right)\)
\(\dfrac{9}{2}\)
Answer: \(\dfrac{9}{2}\)
\(e^x\ge 1\) on \([0,1]\).
\(\int_0^1 (e^x-1)\,dx\)
\(\left[e^x-x\right]_0^1\)
\((e-1)-1\)
\(e-2\)
Answer: \(e-2\)
\(\sin y\ge 0\) on \([0,\pi]\), so the sine curve is to the right of the y-axis.
\(\int_0^{\pi}\sin y\,dy\)
Answer: \(\int_0^{\pi}\sin y\,dy\)
\(2x^2=2\), so \(x=-1\) and \(x=1\).
At \(x=0\), \(1-x^2=1\) is above \(x^2-1=-1\).
\(\int_{-1}^{1}\left((1-x^2)-(x^2-1)\right)\,dx\)
\(\int_{-1}^{1}(2-2x^2)\,dx\)
\(2\int_0^1 (2-2x^2)\,dx\)
\(2\left[2x-\dfrac{2x^3}{3}\right]_0^1\)
\(2\left(2-\dfrac{2}{3}\right)\)
\(\dfrac{8}{3}\)
Answer: \(\dfrac{8}{3}\)
The integrand is bottom minus top, so the integral is the opposite of the area.
The area is \(\dfrac{4}{3}\).
Answer: \(\dfrac{4}{3}\)
Split where the curves trade places.
They meet at \(x=-1,0,1\).
On \([-1,0]\), \(x^3\ge x\).
\(\int_{-1}^{0}(x^3-x)\,dx=\left[\dfrac{x^4}{4}-\dfrac{x^2}{2}\right]_{-1}^{0}\)
\(0-\left(\dfrac{1}{4}-\dfrac{1}{2}\right)\)
\(\dfrac{1}{4}\)
On \([0,1]\), \(x\ge x^3\).
\(\int_0^1 (x-x^3)\,dx=\dfrac{1}{4}\)
Total area: \(\dfrac{1}{2}\).
The single integral \(\int_{-1}^{1}(x-x^3)\,dx\) is 0 because the signed pieces cancel.
Answer: \(\dfrac{1}{2}\)
When \(y=\sqrt{x}\), \(x=y^2\). \(y\) runs from 0 to 3.
The right boundary is \(x=9\) and the left boundary is \(x=y^2\).
\(\int_0^3 (9-y^2)\,dy\)
\(\left[9y-\dfrac{y^3}{3}\right]_0^3\)
\(27-9\)
\(18\)
Answer: \(18\)
\(\int_0^9 \sqrt{x}\,dx=\left[\dfrac{2}{3}x^{3/2}\right]_0^9\)
\(\dfrac{2}{3}\cdot 27\)
\(18\)
Answer: \(18\)
On \([0,\pi/4]\), \(\cos x\ge \sin x\).
\(\int_0^{\pi/4}(\cos x-\sin x)\,dx\)
\(\left[\sin x+\cos x\right]_0^{\pi/4}\)
\(\left(\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}\right)-(0+1)\)
\(\sqrt{2}-1\)
Answer: \(\sqrt{2}-1\)
From \(x=0\) to \(x=3\), \(2x\) is above \(x\).
\(\int_0^3 (2x-x)\,dx\)
\(\int_0^3 x\,dx\)
\(\dfrac{9}{2}\)
Answer: \(\dfrac{9}{2}\)
The curves trade places at \(x=0\).
Signed area cancels. Geometric area adds the absolute pieces.
The area is \(\dfrac{1}{2}\).
Answer: the signed pieces cancel
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