Question: Differentiate $(x-1)\sqrt{4x+1}$. Hence evaluate $\displaystyle\int_2^6 \dfrac{x}{\sqrt{4x+1}}\,dx$.

Step 1. From an earlier sub-group: $\dfrac{d}{dx}\left[(x-1)\sqrt{4x+1}\right] = \dfrac{6x-1}{\sqrt{4x+1}}$.

Step 2. Manipulate the target integrand to match:

$$\dfrac{x}{\sqrt{4x+1}} = \dfrac{1}{6} \cdot \dfrac{6x - 1 + 1}{\sqrt{4x+1}} = \dfrac{1}{6}\left[\dfrac{6x-1}{\sqrt{4x+1}} + \dfrac{1}{\sqrt{4x+1}}\right]$$

Step 3. Integrate from $2$ to $6$:

$$\int_2^6 \dfrac{x}{\sqrt{4x+1}}\,dx = \dfrac{1}{6}\left[(x-1)\sqrt{4x+1}\right]_2^6 + \dfrac{1}{6}\int_2^6 \dfrac{1}{\sqrt{4x+1}}\,dx$$

First piece: $\dfrac{1}{6}\left[5\sqrt{25} - 1\sqrt{9}\right] = \dfrac{1}{6}(25 - 3) = \dfrac{22}{6} = \dfrac{11}{3}$.

Second piece: $\int \dfrac{1}{\sqrt{4x+1}}\,dx = \dfrac{1}{2}\sqrt{4x+1}$. Evaluate: $\dfrac{1}{6} \cdot \dfrac{1}{2}\left[\sqrt{25} - \sqrt{9}\right] = \dfrac{1}{12}(5 - 3) = \dfrac{1}{6}$.

Step 4. Combine: $\dfrac{11}{3} + \dfrac{1}{6} = \dfrac{22 + 1}{6} = \dfrac{23}{6}$.

Answer: $\displaystyle\int_2^6 \dfrac{x}{\sqrt{4x+1}}\,dx = \dfrac{23}{6}$.

Tip: "Differentiate ... hence integrate" requires you to manipulate the target to MATCH the derivative result, often by adding/subtracting a constant to the numerator.

Question: Given $\displaystyle\int_1^6 f(x)\,dx = 14$ and $\displaystyle\int_1^3 f(x)\,dx = 8$, find $\displaystyle\int_1^6 [4x - 3f(x)]\,dx$.

Step 1. Apply linearity:

$$\int_1^6 [4x - 3f(x)]\,dx = \int_1^6 4x\,dx - 3\int_1^6 f(x)\,dx$$

Step 2. First piece: $\displaystyle\int_1^6 4x\,dx = \left[2x^2\right]_1^6 = 72 - 2 = 70$.

Step 3. Second piece is given: $\displaystyle\int_1^6 f(x)\,dx = 14$.

Step 4. Combine:

$$\int_1^6 [4x - 3f(x)]\,dx = 70 - 3 \times 14 = 70 - 42 = 28$$

Answer: $28$.

Tip: Definite-integral algebra rules:

• Linearity: $\int (af + bg) = a\int f + b\int g$
• Additivity: $\int_a^b + \int_b^c = \int_a^c$
• Reversal: $\int_a^b f = -\int_b^a f$
• $\int_a^a f = 0$

Question: Evaluate $\displaystyle\int_0^1 \dfrac{3x^2 + 3x + 3}{(x+2)(x-1)^2}\,dx$. Wait — the limit $x = 1$ would make the denominator zero. Re-state with limits $0$ to $\tfrac{1}{2}$.

Step 1. Partial fractions: $\dfrac{3x^2 + 3x + 3}{(x+2)(x-1)^2} = \dfrac{A}{x+2} + \dfrac{B}{x-1} + \dfrac{C}{(x-1)^2}$.

Multiply: $3x^2 + 3x + 3 = A(x-1)^2 + B(x+2)(x-1) + C(x+2)$.

Cover-up:

• $x = 1$: $3 + 3 + 3 = 9 = C(3) \Rightarrow C = 3$.
• $x = -2$: $12 - 6 + 3 = 9 = A(9) \Rightarrow A = 1$.
• Coeff of $x^2$: $3 = A + B \Rightarrow B = 2$.

Step 2. Integrate:

$$\int \left[\dfrac{1}{x+2} + \dfrac{2}{x-1} + \dfrac{3}{(x-1)^2}\right]dx = \ln|x+2| + 2\ln|x-1| - \dfrac{3}{x-1} + C$$

Step 3. Evaluate from $0$ to $\tfrac{1}{2}$: At $x = \tfrac{1}{2}$: $\ln\tfrac{5}{2} + 2\ln\tfrac{1}{2} - \dfrac{3}{-1/2} = \ln\tfrac{5}{2} - 2\ln 2 + 6$. At $x = 0$: $\ln 2 + 2\ln 1 - \dfrac{3}{-1} = \ln 2 + 3$.

Difference: $\ln\tfrac{5}{2} - 2\ln 2 + 6 - \ln 2 - 3 = \ln\tfrac{5}{2} - 3\ln 2 + 3 = \ln\dfrac{5/2}{8} + 3 = \ln\dfrac{5}{16} + 3$.

Tip: Always do the partial-fraction decomposition first, then integrate each piece, then evaluate. Repeated factor $(x-1)^2 \to$ second term is a power, not a log.

Question: Without computing, argue why $\displaystyle\int_{\pi}^{5\pi/4} \cos x\,dx$ is negative.

Step 1. Examine the integrand on the interval. For $x \in [\pi, 5\pi/4]$:

• At $x = \pi$: $\cos\pi = -1$.
• At $x = 5\pi/4$: $\cos(5\pi/4) = -\dfrac{\sqrt{2}}{2}$.

Across the whole interval, $\cos x < 0$ (the cosine curve lies below the $x$-axis).

Step 2. A definite integral equals the SIGNED area between the curve and the $x$-axis. Since the curve is entirely below the axis on $[\pi, 5\pi/4]$, the signed area is negative.

Verification: $\int \cos x\,dx = \sin x$. At limits: $\sin(5\pi/4) - \sin\pi = -\dfrac{\sqrt{2}}{2} - 0 = -\dfrac{\sqrt{2}}{2} < 0$. $\checkmark$

Tip: Geometric reasoning:

• Integrand $> 0$ on $[a, b]$ $\Rightarrow$ integral $> 0$.
• Integrand $< 0$ on $[a, b]$ $\Rightarrow$ integral $< 0$.
• Integrand changes sign $\Rightarrow$ split at zeros and sum signed areas.

Useful even without computing — saves time on multi-part questions.

Question: Solve for $m$: $\displaystyle\int_0^m \left[2e^{2x} - \dfrac{5}{2}e^{-2x}\right]dx = \dfrac{3}{4}$.

Step 1. Integrate:

$$\int \left[2e^{2x} - \dfrac{5}{2}e^{-2x}\right]dx = e^{2x} + \dfrac{5}{4}e^{-2x} + C$$

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

$$\left[e^{2x} + \dfrac{5}{4}e^{-2x}\right]_0^m = \left(e^{2m} + \dfrac{5}{4}e^{-2m}\right) - \left(1 + \dfrac{5}{4}\right) = e^{2m} + \dfrac{5}{4}e^{-2m} - \dfrac{9}{4}$$

Step 3. Set equal to $\dfrac{3}{4}$:

$$e^{2m} + \dfrac{5}{4}e^{-2m} = 3$$

Multiply through by $4e^{2m}$:

$$4e^{4m} + 5 = 12e^{2m} \Rightarrow 4e^{4m} - 12e^{2m} + 5 = 0$$

Step 4. Let $u = e^{2m}$. Quadratic: $4u^2 - 12u + 5 = 0$.

$$u = \dfrac{12 \pm \sqrt{144 - 80}}{8} = \dfrac{12 \pm 8}{8} = \dfrac{5}{2} \text{ or } \dfrac{1}{2}$$

So $e^{2m} = \dfrac{5}{2}$ or $\dfrac{1}{2}$, giving $m = \dfrac{1}{2}\ln\dfrac{5}{2}$ or $m = \dfrac{1}{2}\ln\dfrac{1}{2} = -\dfrac{1}{2}\ln 2$.

Tip: Definite-integral-equals-constant problems: integrate, evaluate, equate, solve. Substitution $u = e^{2m}$ often turns exponentials into a quadratic.

Question: Evaluate $\displaystyle\int_0^{\pi/4} 2\sin^2\theta\,d\theta$ using a double-angle identity.

Step 1. Apply identity $2\sin^2\theta = 1 - \cos 2\theta$:

$$\int_0^{\pi/4} 2\sin^2\theta\,d\theta = \int_0^{\pi/4} (1 - \cos 2\theta)\,d\theta$$

Step 2. Integrate term-by-term:

$$\left[\theta - \dfrac{1}{2}\sin 2\theta\right]_0^{\pi/4}$$

Step 3. Evaluate:

• At $\theta = \pi/4$: $\dfrac{\pi}{4} - \dfrac{1}{2}\sin\dfrac{\pi}{2} = \dfrac{\pi}{4} - \dfrac{1}{2}$
• At $\theta = 0$: $0 - 0 = 0$

Difference: $\dfrac{\pi}{4} - \dfrac{1}{2}$.

Answer: $\dfrac{\pi}{4} - \dfrac{1}{2} \approx 0.285$.

Tip: For integrands like $\sin^2 x$ or $\cos^2 x$ that aren't directly integrable, use double-angle identities first. The integrand becomes a constant plus a $\cos 2\theta$-like term, both directly integrable.