Question: Curve $y = \dfrac{10}{4x+1}$, tangent at $P(1, 2)$. Find the area bounded by the curve, the tangent, and the $x$-axis from $x = 0$ to $x = 1$.

Step 1. Differentiate to find tangent gradient:

$$\dfrac{dy}{dx} = -\dfrac{40}{(4x+1)^2}, \quad \text{at } x=1: \dfrac{dy}{dx} = -\dfrac{40}{25} = -\dfrac{8}{5}$$

Step 2. Tangent equation: $y - 2 = -\dfrac{8}{5}(x - 1) \Rightarrow y = -\dfrac{8}{5}x + \dfrac{18}{5}$.

$x$-intercept: $0 = -\dfrac{8}{5}x + \dfrac{18}{5} \Rightarrow x = \dfrac{18}{8} = \dfrac{9}{4}$.

Step 3. Area under curve from $0$ to $1$:

$$\int_0^1 \dfrac{10}{4x+1}\,dx = \dfrac{10}{4}\left[\ln|4x+1|\right]_0^1 = \dfrac{5}{2}\ln 5$$

Step 4. Area of triangle under tangent from $x = 1$ to $x = 9/4$ (above $x$-axis): base $= \dfrac{9}{4} - 1 = \dfrac{5}{4}$, height $= 2$.

$$\text{Triangle area} = \dfrac{1}{2} \times \dfrac{5}{4} \times 2 = \dfrac{5}{4}$$

Step 5. Total bounded area $= \dfrac{5}{2}\ln 5 + \dfrac{5}{4} \approx 4.02 + 1.25 \approx 5.27$ square units.

Tip: Combo problems (curve + line + axis): split the region into an integral piece + a geometric piece (triangle/trapezium). Sum them up.

Question: Curve $y = \dfrac{8}{2+3x}$ intersects the line $y = -\dfrac{3x}{2} + 4$ at $x = 0$ and $x = 2$. Find the area enclosed.

Step 1. Verify intersections:

• $x = 0$: curve $y = 4$; line $y = 4$. $\checkmark$
• $x = 2$: curve $y = \dfrac{8}{8} = 1$; line $y = -3 + 4 = 1$. $\checkmark$

Step 2. Identify which is "upper". Test $x = 1$:

• Curve: $y = \dfrac{8}{5} = 1.6$
• Line: $y = -1.5 + 4 = 2.5$ — higher.

So line > curve on $(0, 2)$.

Step 3. Enclosed area:

$$\text{Area} = \int_0^2 [\text{line} - \text{curve}]\,dx = \int_0^2 \left[-\dfrac{3x}{2} + 4 - \dfrac{8}{2+3x}\right]dx$$

Step 4. Antiderivative:

$$-\dfrac{3x^2}{4} + 4x - \dfrac{8}{3}\ln|2+3x|$$

Step 5. Evaluate from $0$ to $2$:

$$\left[-3 + 8 - \dfrac{8}{3}\ln 8\right] - \left[0 + 0 - \dfrac{8}{3}\ln 2\right] = 5 - \dfrac{8}{3}(\ln 8 - \ln 2) = 5 - \dfrac{8}{3}\ln 4 = 5 - \dfrac{16}{3}\ln 2$$

Numerically: $\approx 5 - 3.696 \approx 1.30$ square units.

Tip: Area between two curves $= \int_a^b (\text{upper} - \text{lower})\,dx$. Always check which is upper by plugging a test value.

Question: Find the total area enclosed between the curve $y = x^3 - x$ and the $x$-axis.

Step 1. Find roots: $x^3 - x = x(x^2 - 1) = x(x-1)(x+1) = 0 \Rightarrow x = -1, 0, 1$.

Step 2. Determine sign of $y$ on each interval:

• $(-1, 0)$: at $x = -0.5$: $y = -0.125 + 0.5 = 0.375 > 0$ (above axis).
• $(0, 1)$: at $x = 0.5$: $y = 0.125 - 0.5 = -0.375 < 0$ (below axis).

Step 3. Compute the two pieces separately (use absolute value for the negative one):

$$A_1 = \int_{-1}^0 (x^3 - x)\,dx = \left[\dfrac{x^4}{4} - \dfrac{x^2}{2}\right]_{-1}^0 = 0 - \left(\dfrac{1}{4} - \dfrac{1}{2}\right) = \dfrac{1}{4}$$ $$A_2 = \left|\int_0^1 (x^3 - x)\,dx\right| = \left|\left[\dfrac{x^4}{4} - \dfrac{x^2}{2}\right]_0^1\right| = \left|\dfrac{1}{4} - \dfrac{1}{2}\right| = \dfrac{1}{4}$$

Step 4. Total area: $A_1 + A_2 = \dfrac{1}{2}$ square units.

Tip: If $y$ changes sign in the integration range, the unmodified definite integral gives the net signed area (not the geometric area). Split at zeros and add absolute values for total area.

Question: Find the area bounded by the curve $x = (y - 2)^2$, the line $y = 4$, and the $y$-axis (in the first quadrant).

Step 1. Note the curve is given as $x = f(y)$. Integrate w.r.t. $y$ to get the area between the curve and the $y$-axis.

Step 2. When does the curve meet $y$-axis ($x = 0$)? $(y-2)^2 = 0 \Rightarrow y = 2$. Region spans $y = 2$ to $y = 4$.

Step 3. Area:

$$\text{Area} = \int_2^4 x\,dy = \int_2^4 (y-2)^2\,dy$$

Step 4. Antiderivative:

$$\int (y-2)^2\,dy = \dfrac{(y-2)^3}{3}$$

Step 5. Evaluate:

$$\left[\dfrac{(y-2)^3}{3}\right]_2^4 = \dfrac{8}{3} - 0 = \dfrac{8}{3}$$

Answer: $\text{Area} = \dfrac{8}{3}$ square units.

Tip: Curves of the form $x = f(y)$ (sideways parabolas, inverse functions) are easier to integrate w.r.t. $y$ than to rearrange. Just check the bounds in $y$ (not $x$).

Question: The curve $y = \sin 2x + x$ has a minimum point $M$ at $x = \dfrac{2\pi}{3}$. The chord $OM$ connects the origin to $M$. Find the area bounded by the curve and the chord from $x = 0$ to $x = \dfrac{2\pi}{3}$.

Step 1. At $x = \dfrac{2\pi}{3}$: $y = \sin\dfrac{4\pi}{3} + \dfrac{2\pi}{3} = -\dfrac{\sqrt{3}}{2} + \dfrac{2\pi}{3}$.

So $M = \left(\dfrac{2\pi}{3}, \dfrac{2\pi}{3} - \dfrac{\sqrt{3}}{2}\right)$.

Step 2. Chord $OM$ has gradient $= \dfrac{2\pi/3 - \sqrt{3}/2}{2\pi/3} = 1 - \dfrac{3\sqrt{3}}{4\pi}$. Equation: $y = mx$ where $m = 1 - \dfrac{3\sqrt{3}}{4\pi}$.

Step 3. Area between chord (above) and curve (below) — assuming chord is above curve on this interval:

$$\text{Area} = \int_0^{2\pi/3} (mx - \sin 2x - x)\,dx = \int_0^{2\pi/3} \left[(m-1)x - \sin 2x\right]dx$$

Step 4. Antiderivative:

$$\dfrac{(m-1)x^2}{2} + \dfrac{1}{2}\cos 2x$$

At $x = 2\pi/3$: $\dfrac{(m-1)(2\pi/3)^2}{2} + \dfrac{1}{2}\cos\dfrac{4\pi}{3} = \dfrac{(m-1) \cdot 4\pi^2/9}{2} - \dfrac{1}{4}$

At $x = 0$: $0 + \dfrac{1}{2} = \dfrac{1}{2}$.

Substituting $m - 1 = -\dfrac{3\sqrt{3}}{4\pi}$ gives $\dfrac{-3\sqrt{3}/(4\pi) \cdot 4\pi^2/9}{2} - \dfrac{1}{4} - \dfrac{1}{2} = -\dfrac{\sqrt{3}\pi}{6} - \dfrac{3}{4}$.

Take absolute value: $\dfrac{\sqrt{3}\pi}{6} + \dfrac{3}{4}$.

Tip: Curve-and-chord area: $|\int(\text{chord} - \text{curve})\,dx|$ over the interval bounded by the chord endpoints. The chord equation is $y = mx + c$ through the two endpoints.

Question: Show that $\dfrac{d}{dx}[x^2 \ln x - x] = 2x\ln x + x - 1$. Hence find the area enclosed between $y = 2x\ln x$, the $x$-axis, and $x = 2$ (for $x \geq 1$).

Step 1. Differentiate by product rule:

$$\dfrac{d}{dx}[x^2 \ln x] = 2x\ln x + x^2 \cdot \dfrac{1}{x} = 2x\ln x + x$$

So $\dfrac{d}{dx}[x^2\ln x - x] = 2x\ln x + x - 1$. $\checkmark$

Step 2. Integrate: $\int (2x\ln x + x - 1)\,dx = x^2\ln x - x + C$.

So $\int 2x\ln x\,dx = x^2\ln x - x - \int(x - 1)\,dx + C = x^2\ln x - x - \dfrac{x^2}{2} + x + C = x^2\ln x - \dfrac{x^2}{2} + C$.

Step 3. Area from $x = 1$ to $x = 2$ (curve crosses $x$-axis at $x = 1$ since $\ln 1 = 0$):

$$\int_1^2 2x\ln x\,dx = \left[x^2\ln x - \dfrac{x^2}{2}\right]_1^2 = \left[4\ln 2 - 2\right] - \left[0 - \dfrac{1}{2}\right] = 4\ln 2 - \dfrac{3}{2}$$

Numerically: $\approx 2.77 - 1.5 = 1.27$ square units.

Tip: "Show derivative ... hence area": the integration step usually requires you to subtract the easy-to-integrate terms from the hard one, then solve for the wanted integral.

Question: Show that $\dfrac{7}{2}\ln 2 < \displaystyle\int_{-1/3}^2 \ln(3x+2)\,dx < 7\ln 2$ by sandwiching the integrand.

Step 1. The integrand $\ln(3x+2)$ on $x \in [-\tfrac{1}{3}, 2]$:

• At $x = -1/3$: $\ln 1 = 0$ (left endpoint).
• At $x = 2$: $\ln 8 = 3\ln 2$ (right endpoint).

Since $\ln$ is increasing, the integrand increases from $0$ to $3\ln 2$ across the interval.

Step 2. Lower bound — the integrand is BIGGER than the chord from $(−1/3, 0)$ to $(2, 3\ln 2)$ at each point (since $\ln$ is concave, the function lies ABOVE its chord). The chord triangle has base $2 - (-\tfrac{1}{3}) = \tfrac{7}{3}$ and height $3\ln 2$:

$$\text{Chord triangle area} = \dfrac{1}{2} \cdot \dfrac{7}{3} \cdot 3\ln 2 = \dfrac{7}{2}\ln 2$$

So $\displaystyle\int_{-1/3}^2 \ln(3x+2)\,dx > \dfrac{7}{2}\ln 2$.

Step 3. Upper bound — circumscribed rectangle with width $\dfrac{7}{3}$ and height $3\ln 2$ (the max value):

$$\text{Rectangle area} = \dfrac{7}{3} \cdot 3\ln 2 = 7\ln 2$$

So $\displaystyle\int_{-1/3}^2 \ln(3x+2)\,dx < 7\ln 2$.

Step 4. Combined: $\dfrac{7}{2}\ln 2 < \displaystyle\int_{-1/3}^2 \ln(3x+2)\,dx < 7\ln 2$. $\checkmark$

Tip: Geometric inequality bounds:

• Lower bound: inscribed rectangle OR chord triangle (for concave functions).
• Upper bound: circumscribed rectangle (max value × width).

No exact area calculation needed.