Question: Given $\dfrac{d^2y}{dx^2} = 6x - 2$, the curve passes through $(2, -9)$ with gradient $3$ at that point. Find $y$.

Step 1. First antiderivative:

$$\dfrac{dy}{dx} = \int (6x - 2)\,dx = 3x^2 - 2x + C_1$$

Step 2. Apply gradient condition at $x = 2$:

$$3 = 3(4) - 2(2) + C_1 = 12 - 4 + C_1 = 8 + C_1 \Rightarrow C_1 = -5$$

So $\dfrac{dy}{dx} = 3x^2 - 2x - 5$.

Step 3. Second antiderivative:

$$y = \int (3x^2 - 2x - 5)\,dx = x^3 - x^2 - 5x + C_2$$

Step 4. Apply point condition $(2, -9)$:

$$-9 = 8 - 4 - 10 + C_2 = -6 + C_2 \Rightarrow C_2 = -3$$

Answer: $y = x^3 - x^2 - 5x - 3$.

Tip: Two antiderivative steps = two constants of integration. Pin them down in order: gradient condition fixes $C_1$, point condition fixes $C_2$.

Question: $\dfrac{d^2y}{dx^2} = (\sin x - \cos x)^2$. Find $y$ given that the curve passes through $(0, 0)$ with gradient $-1$ at that point.

Step 1. Expand the integrand using Pythagorean identity:

$$(\sin x - \cos x)^2 = \sin^2 x - 2\sin x \cos x + \cos^2 x = 1 - \sin 2x$$

Step 2. First antiderivative:

$$\dfrac{dy}{dx} = \int (1 - \sin 2x)\,dx = x + \dfrac{1}{2}\cos 2x + C_1$$

Apply gradient condition at $x = 0$:

$$-1 = 0 + \dfrac{1}{2}\cos 0 + C_1 = \dfrac{1}{2} + C_1 \Rightarrow C_1 = -\dfrac{3}{2}$$

Step 3. Second antiderivative:

$$y = \int \left(x + \dfrac{1}{2}\cos 2x - \dfrac{3}{2}\right)dx = \dfrac{x^2}{2} + \dfrac{1}{4}\sin 2x - \dfrac{3x}{2} + C_2$$

Apply $(0, 0)$: $0 = 0 + 0 - 0 + C_2 \Rightarrow C_2 = 0$.

Answer: $y = \dfrac{x^2}{2} + \dfrac{1}{4}\sin 2x - \dfrac{3x}{2}$.

Tip: ALWAYS simplify trig integrands using identities BEFORE integrating. $\sin^2 x \to \dfrac{1 - \cos 2x}{2}$, $\cos^2 x \to \dfrac{1 + \cos 2x}{2}$, $\sin x \cos x \to \dfrac{1}{2}\sin 2x$.

Question: $\dfrac{d^2y}{dx^2} = 3e^{-x} + 8e^{2x}$. Find $y$ given $\dfrac{dy}{dx} = 0$ when $x = 0$ and $y = 0$ when $x = 0$.

Step 1. First antiderivative:

$$\dfrac{dy}{dx} = \int (3e^{-x} + 8e^{2x})\,dx = -3e^{-x} + 4e^{2x} + C_1$$

Apply gradient condition at $x = 0$:

$$0 = -3 + 4 + C_1 = 1 + C_1 \Rightarrow C_1 = -1$$

Step 2. Second antiderivative:

$$y = \int (-3e^{-x} + 4e^{2x} - 1)\,dx = 3e^{-x} + 2e^{2x} - x + C_2$$

Apply $(0, 0)$:

$$0 = 3 + 2 - 0 + C_2 = 5 + C_2 \Rightarrow C_2 = -5$$

Answer: $y = 3e^{-x} + 2e^{2x} - x - 5$.

Tip: $\int e^{ax}\,dx = \dfrac{1}{a}e^{ax} + C$. The factor of $\dfrac{1}{a}$ is the most-forgotten step. Pay attention especially to negative exponents.

Question: $\dfrac{d^2y}{dx^2} = \dfrac{8}{(4x+1)^2}$. Find $y$ given gradient $-1$ at $x = 0$ and $y = 2$ at $x = 0$.

Step 1. First antiderivative:

$$\dfrac{dy}{dx} = \int 8(4x+1)^{-2}\,dx = 8 \cdot \dfrac{(4x+1)^{-1}}{4 \cdot (-1)} = -\dfrac{2}{4x+1} + C_1$$

Apply gradient at $x = 0$:

$$-1 = -\dfrac{2}{1} + C_1 = -2 + C_1 \Rightarrow C_1 = 1$$

Step 2. Second antiderivative:

$$y = \int \left[-\dfrac{2}{4x+1} + 1\right]dx = -\dfrac{2}{4}\ln|4x+1| + x + C_2 = -\dfrac{1}{2}\ln|4x+1| + x + C_2$$

Apply $(0, 2)$:

$$2 = -\dfrac{1}{2}\ln 1 + 0 + C_2 = C_2 \Rightarrow C_2 = 2$$

Answer: $y = -\dfrac{1}{2}\ln|4x+1| + x + 2$.

Tip: $\int \dfrac{1}{(ax+b)^n}\,dx$ gives:

• A log if $n = 1$: $\dfrac{1}{a}\ln|ax+b|$
• A power if $n \neq 1$: $\dfrac{(ax+b)^{1-n}}{a(1-n)}$

Watch the sign carefully: $n = 2$ gives $\dfrac{-1}{a(ax+b)}$.

Question: Show $\dfrac{d}{dx}\left[\dfrac{\ln x}{x^4}\right] = \dfrac{1 - 4\ln x}{x^5}$. Hence find $\displaystyle\int \dfrac{\ln x}{x^5}\,dx$.

Step 1. Differentiate by quotient rule. Let $u = \ln x$, $v = x^4$:

• $u' = 1/x$, $v' = 4x^3$

$$\dfrac{d}{dx}\left[\dfrac{\ln x}{x^4}\right] = \dfrac{(1/x)(x^4) - (\ln x)(4x^3)}{x^8} = \dfrac{x^3 - 4x^3 \ln x}{x^8} = \dfrac{1 - 4\ln x}{x^5} \;\checkmark$$

Step 2. Integrate both sides:

$$\int \dfrac{1 - 4\ln x}{x^5}\,dx = \dfrac{\ln x}{x^4} + C$$

Step 3. Split:

$$\int \dfrac{1}{x^5}\,dx - 4\int \dfrac{\ln x}{x^5}\,dx = \dfrac{\ln x}{x^4} + C$$ $$-\dfrac{1}{4x^4} - 4\int \dfrac{\ln x}{x^5}\,dx = \dfrac{\ln x}{x^4} + C$$

Step 4. Solve:

$$4\int \dfrac{\ln x}{x^5}\,dx = -\dfrac{1}{4x^4} - \dfrac{\ln x}{x^4} - C$$ $$\int \dfrac{\ln x}{x^5}\,dx = -\dfrac{1}{16 x^4} - \dfrac{\ln x}{4x^4} + C$$

Tip: "Show ... hence find" works for $\int \dfrac{\ln x}{x^n}\,dx$-style integrals. Differentiate the cued expression, then algebraically rearrange to isolate the target integral.

Question: $\dfrac{d^2y}{dx^2} = 3k(\sqrt{x} - 1)$. The curve has a stationary point at $x = 4$ and passes through $(0, 1)$. Given the curve also passes through $(4, -5)$, find $k$ and the equation of the curve.

Step 1. First antiderivative:

$$\dfrac{dy}{dx} = \int 3k(x^{1/2} - 1)\,dx = 3k \cdot \dfrac{2}{3}x^{3/2} - 3k x + C_1 = 2k x^{3/2} - 3k x + C_1$$

Stationary point at $x = 4$ means $\dfrac{dy}{dx} = 0$ there:

$$0 = 2k(8) - 3k(4) + C_1 = 16k - 12k + C_1 = 4k + C_1 \Rightarrow C_1 = -4k$$

Step 2. Second antiderivative:

$$y = \int (2k x^{3/2} - 3k x - 4k)\,dx = \dfrac{4k}{5}x^{5/2} - \dfrac{3k}{2}x^2 - 4k x + C_2$$

Apply $(0, 1)$: $1 = C_2$.

Apply $(4, -5)$:

$$-5 = \dfrac{4k}{5}(32) - \dfrac{3k}{2}(16) - 4k(4) + 1 = \dfrac{128k}{5} - 24k - 16k + 1 = \dfrac{128k - 200k}{5} + 1 = -\dfrac{72k}{5} + 1$$ $$-6 = -\dfrac{72k}{5} \Rightarrow k = \dfrac{30}{72} = \dfrac{5}{12}$$

Answer: $k = \dfrac{5}{12}$, then sub back to get $y$.

Tip: When the integrand has an unknown parameter, you need an EXTRA condition (beyond what's needed to pin the integration constants). Count carefully.

Question: $\dfrac{dy}{dx} = 3x^2 - 6x + 8$. The curve passes through $(1, 6)$. (a) Find $y$. (b) Show that the gradient is never less than $5$.

Step 1. Antiderivative:

$$y = \int (3x^2 - 6x + 8)\,dx = x^3 - 3x^2 + 8x + C$$

Apply $(1, 6)$: $6 = 1 - 3 + 8 + C = 6 + C \Rightarrow C = 0$.

Answer (a): $y = x^3 - 3x^2 + 8x$.

Step 2 (b). Gradient $= \dfrac{dy}{dx} = 3x^2 - 6x + 8$. To show this $\geq 5$ always, complete the square:

$$3x^2 - 6x + 8 = 3(x^2 - 2x) + 8 = 3[(x-1)^2 - 1] + 8 = 3(x-1)^2 + 5$$

Step 3. Since $(x-1)^2 \geq 0$, $3(x-1)^2 \geq 0$, hence:

$$\dfrac{dy}{dx} = 3(x-1)^2 + 5 \geq 5$$

Minimum gradient $= 5$, occurring at $x = 1$. $\checkmark$

Tip: "Show gradient $\geq k$" usually reduces to completing the square on $\dfrac{dy}{dx}$. The vertex of $a(x-h)^2 + k$ gives the min (if $a > 0$) or max (if $a < 0$).

Question: The curve $y = \sqrt{3x+4}$ meets the $x$-axis when... no — $y \geq 2$ everywhere. Re-state: Find the area enclosed by the curve $y = \sqrt{3x+4}$, the $x$-axis, the $y$-axis, and the line $x = 4$.

Step 1. $y$-intercept: at $x = 0$, $y = 2$. At $x = 4$: $y = 4$. Region is the area under the curve from $x = 0$ to $x = 4$.

Step 2. Set up the definite integral:

$$\text{Area} = \int_0^4 \sqrt{3x+4}\,dx = \int_0^4 (3x+4)^{1/2}\,dx$$

Step 3. Antiderivative:

$$\int (3x+4)^{1/2}\,dx = \dfrac{(3x+4)^{3/2}}{3 \cdot (3/2)} = \dfrac{2(3x+4)^{3/2}}{9}$$

Step 4. Evaluate:

$$\dfrac{2}{9}\left[(16)^{3/2} - (4)^{3/2}\right] = \dfrac{2}{9}[64 - 8] = \dfrac{2 \times 56}{9} = \dfrac{112}{9}$$

Answer: $\text{Area} = \dfrac{112}{9} \approx 12.44$ square units.

Tip: "Area under curve from $a$ to $b$" $= \int_a^b y\,dx$. For surd integrands $\sqrt{ax+b}$, rewrite as $(ax+b)^{1/2}$ and apply the inner-coefficient rule: divide by $a$ AND by $n+1 = 3/2$.