Area Under THE Curve flash cards

1. 1.Give the formula for the area bounded by the curve $y=f(x)$, the $x$-axis, and the ordinates $x=a$ and $x=b$ (where $f(x)\ge 0$ on $[a,b]$).

   $A=\int_a^b y\,dx=\int_a^b f(x)\,dx$

   Hint: Think vertical strips of width $dx$.

2. 2.How is the area between two curves $y=f(x)$ and $y=g(x)$ (with $f(x)\ge g(x)$) from $x=a$ to $x=b$ computed?

   $A=\int_a^b \big[f(x)-g(x)\big]\,dx$

   Hint: Upper curve minus lower curve.

3. 3.What is the standard formula for area with respect to the $y$-axis, bounded by $x=f(y)$ and the $y$-axis between $y=c$ and $y=d$?

   $A=\int_c^d x\,dy=\int_c^d f(y)\,dy$

   Hint: Use horizontal strips of width $dy$.

4. 4.How do you find the area of a region when the curve lies below the $x$-axis over part of $[a,b]$?

   Take the modulus of the integral over each part separately: $A=\left|\int_a^c y\,dx\right|+\left|\int_c^b y\,dx\right|$, splitting at points where $y=0$.

   Hint: Area is always taken as positive; find the zeros first.

5. 5.State the area enclosed by the circle $x^2+y^2=a^2$ and the area of the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.

   Circle area $=\pi a^2$; Ellipse area $=\pi a b$.

   Hint: Both are standard results, derivable via integration using symmetry ($4\times$ first quadrant area).

6. 6.Find the total area enclosed between $y=x^3-3x$ and $y=x$.

   Setting $x^3-3x=x$ gives $x^3-4x=x(x-2)(x+2)=0$, so the curves meet at $x=-2,0,2$. Since $x^3-4x$ is an odd function, the two lobes have equal area, so total area $=2\int_0^2(4x-x^3)\,dx=2\left[2x^2-\dfrac{x^4}{4}\right]_0^2=2(8-4)=8$.

   Hint: Locate all three roots first, then use oddness of $x^3-4x$ to double one lobe.

7. 7.The curves $y=x^4-2x^2$ and $y=2x^2$ touch at one point and cross at two others. Find the total enclosed area.

   $x^4-2x^2=2x^2\Rightarrow x^2(x^2-4)=0$, giving $x=0$ (a point of tangency, not a crossing) and $x=\pm2$. Since $x^2-4<0$ throughout $(-2,2)$, the line $y=2x^2$ stays above the quartic on the whole interval despite the touch at $0$. Area $=\int_{-2}^{2}(4x^2-x^4)\,dx=2\int_0^2(4x^2-x^4)\,dx=2\left[\dfrac{4x^3}{3}-\dfrac{x^5}{5}\right]_0^2=\dfrac{128}{15}$.

   Hint: Check whether $x=0$ is a genuine crossing or just a tangency before splitting the integral.

8. 8.Find the area enclosed between $y=\sin x$ and $y=\cos x$ over a full period $[0,2\pi]$.

   They intersect where $\tan x=1$, i.e. $x=\dfrac{\pi}{4},\dfrac{5\pi}{4}$ in $[0,2\pi)$. Writing $\sin x-\cos x=\sqrt2\sin\left(x-\dfrac{\pi}{4}\right)$, the required area is $\displaystyle\int_0^{2\pi}\sqrt2\left|\sin\left(x-\dfrac{\pi}{4}\right)\right|dx=\sqrt2\int_0^{2\pi}|\sin u|\,du=\sqrt2\cdot4=4\sqrt2$, using $\int_0^{2\pi}|\sin u|\,du=4$.

   Hint: Combine $\sin x-\cos x$ into a single sinusoid before integrating the absolute value.

9. 9.Find the total area enclosed between $y=x^3-6x$ and $y=x$.

   $x^3-6x=x\Rightarrow x^3-7x=x(x^2-7)=0$, so $x=0,\pm\sqrt7$. The function $x^3-7x$ is odd, so total area $=2\int_0^{\sqrt7}(7x-x^3)\,dx=2\left[\dfrac{7x^2}{2}-\dfrac{x^4}{4}\right]_0^{\sqrt7}=2\left(\dfrac{49}{2}-\dfrac{49}{4}\right)=\dfrac{49}{2}$.

   Hint: Find all roots of the odd cubic difference, then double the area on one side.

10. 10.Curves $y=x^3$ and $y=3x^2-2x$ intersect at three points. Find the total area they enclose.

    $x^3=3x^2-2x\Rightarrow x(x-1)(x-2)=0$, so $x=0,1,2$. On $(0,1)$, $y=x^3$ lies above; on $(1,2)$ the parabola lies above (check sign of $x^3-3x^2+2x$ at $x=1.5$). With $H(x)=\dfrac{x^4}{4}-x^3+x^2$: area $=[H(1)-H(0)]-[H(2)-H(1)]=\dfrac14-\left(0-\dfrac14\right)=\dfrac12$.

    Hint: Find the sign of $x^3-3x^2+2x$ separately on $(0,1)$ and $(1,2)$.

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