Calculus Recap (Polytechnic)

Limits

Problem 1. Evaluate the limit $\displaystyle \lim_{x \to 1} (x+1)$ .

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Solution. Substituting $x = 1$ ,

$\displaystyle \begin{aligned} \lim_{x \to 1} (x+1) &= 1 + 1 = 2. \end{aligned}$

Problem 2. Evaluate the limit $\displaystyle \lim_{x \to 1} \frac{x^2 - 1}{x-1}$ .

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Solution. Factorising for $x \neq 1$ ,

$\displaystyle \begin{aligned} \lim_{x \to 1} \frac{x^2 - 1}{x-1} &= \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} \\ &= \lim_{x \to 1} (x+1) = 2, \end{aligned}$

where the last $=$ follows from Problem 1.

Problem 3. Evaluate the limit $\displaystyle \lim_{x \to 1} \frac{\sqrt x - 1}{x-1}$ .

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Solution. Rationalising the numerator then substituting $x = 1$ ,

$\displaystyle \begin{aligned} \lim_{x \to 1} \frac{\sqrt x - 1}{x-1} &= \lim_{x \to 1} \frac{\sqrt x - 1}{x-1} \cdot \frac{\sqrt x + 1}{\sqrt x + 1} \\ &= \lim_{x \to 1} \frac{(\sqrt x)^2 - 1^2}{(x-1)(\sqrt x + 1)} \\ &= \lim_{x \to 1} \frac{x - 1}{(x-1)(\sqrt x + 1)} \\ &= \lim_{x \to 1} \frac 1{\sqrt x + 1} = \frac 1{\sqrt 1 + 1} = \frac 12.\end{aligned}$

Problem 4. Without using differentiation, compute the gradient of the tangent to $y =x^2$ at $(1, 1)$ .

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Solution. Denote $(x_1, y_1) = (1, 1)$ and $(x_2, y_2) = (x, x^2)$ . By definition of the gradient of the tangent,

$\displaystyle \begin{aligned} \lim_{x \to 1} \frac{\text{(rise)}}{\text{(run)}} &= \lim_{x \to 1} \frac{y_2 - y_1}{x_2 - x_1} = \lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2,\end{aligned}$

where the last $=$ follows from Problem 2.

Differentiation

Problem 5. Using differentiation, compute the gradient of the tangent to $y = x^2$ at $(1, 1)$ .

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Solution. Differentiating,

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = \frac{\mathrm d}{\mathrm dx} (x^2) = 2x.$

Setting $x = 1$ , $\displaystyle \frac{\mathrm dy}{\mathrm dx} = 2 \cdot 1 = 2$ .

Problem 6. Defining $f(x) = x^2$ , evaluate $f''(1)$ .

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Solution. Differentiating twice,

$\begin{aligned} f(x) &= x^2, \\ f'(x) &= 2x, \\ f''(x) &= 2. \end{aligned}$

Setting $x = 1$ , $f''(1) = 2$ .

Problem 7. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx}\left(\ln(x^2+9) \right)$ .

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Solution. Employing the chain rule,

$\displaystyle \begin{aligned} \frac{ \mathrm d }{ \mathrm dx }\left( \ln(x^2+1) \right) &= \frac 1{x^2 + 9} \cdot \frac{\mathrm d}{\mathrm dx}(x^2 + 1) \\ &= \frac 1{x^2 + 9} \cdot 2x \\ &= \frac { 2x }{x^2 + 9}. \end{aligned}$

Problem 8. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx}\left(\sin \left(\ln(x^2+9) \right) \right)$ .

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Solution. Employing the chain rule and Problem 7,

$\displaystyle \begin{aligned} \frac{\mathrm d}{\mathrm dx}\left(\sin \left(\ln(x^2+1) \right) \right) &= \cos\left(\ln(x^2+9) \right) \cdot \frac{\mathrm d}{\mathrm dx}\ln(x^2+9) \\ &= \cos\left(\ln(x^2+9) \right) \cdot \frac {2x}{x^2 + 9} \\ &= \frac {2x \cos\left(\ln(x^2+9) \right) }{x^2 + 9}. \end{aligned}$

Problem 9. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} \left( (\ln x) \cdot (x^2 + 9) \right)$ .

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Solution. Employing the product rule,

$\displaystyle \begin{aligned} \frac{\mathrm d}{\mathrm dx} \left( (\ln x) \cdot (x^2 + 9) \right) &= (x^2 + 9) \cdot \frac{\mathrm d}{\mathrm dx} ( \ln x ) + \ln x \cdot \frac{\mathrm d}{\mathrm dx} (x^2 + 9) \\ &= (x^2 + 9) \cdot \frac 1x + \ln x \cdot (2x) \\ &= x + \frac 9x + 2x \ln x. \end{aligned}$

Problem 10. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} \left( \frac{\ln x}{x^2 + 9}\right)$ .

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Solution. Employing the quotient rule,

$\displaystyle \begin{aligned} \frac{\mathrm d}{\mathrm dx} \left( \frac{\ln x}{x^2 + 9} \right) &= \frac{(x^2 +9) \cdot \frac{\mathrm d}{\mathrm dx} ( \ln x ) - \ln x \cdot \frac{\mathrm d}{\mathrm dx} (x^2 + 9) }{(x^2 + 9)^2} \\ &= \frac{ (x^2 + 9) \cdot \frac 1x - \ln x \cdot 2x}{(x^2 + 9)^2 } \\ &= \frac{x^2 + 9 -2x^2 \ln x}{x(x^2 + 9)^2}. \end{aligned}$

Integration

Problem 11. Evaluate $\displaystyle \int \left(2\ln x + \frac 5{x^2+9}\right)\, \mathrm dx$ .

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Solution. Applying the linearity of integration and using common formulas (here and here),

$\displaystyle \begin{aligned} \int \left(2\ln x + \frac 5{x^2+9}\right)\, \mathrm dx &= \int \left(2 \cdot\ln x + 5 \cdot \frac 1{x^2 + 9}\right)\, \mathrm dx \\ &= 2 \cdot \int \ln x \, \mathrm dx + 5 \cdot \int \frac 1{3^2 + x^2} \, \mathrm dx \\ &= 2 \cdot (x \ln x - x) + 5 \cdot \frac 1{3} \tan^{-1} \left(\frac x3 \right) + C \\ &= 2x \ln x - 2x + \frac 53 \tan^{-1} \left(\frac x3 \right) + C. \end{aligned}$

Problem 12. Evaluate $\displaystyle \int_1^2 x\ln (x^2+9)\, \mathrm dx$ .

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Solution. Making the substitution $\displaystyle u = x^2 + 9 \Rightarrow \mathrm du = 2x\, \mathrm dx$ ,

$\displaystyle \begin{aligned} \int x\ln (x^2+9 )\, \mathrm dx &= \frac 12 \int \ln (x^2+9 ) \cdot 2x\, \mathrm dx \\ &= \frac 12 \int \ln u \, \mathrm du \\ &= \frac 12 (u \ln u - u) + C \\ &= \frac 12 \left((x^2 + 9) \ln (x^2 + 9) - (x^2 + 9) \right) + C. \end{aligned}$

Hence, substituting the limits,

$\displaystyle \begin{aligned} \int_1^2 x\ln (x^2+1)\, \mathrm dx &= \left[ \frac 12 \left( (x^2 + 9) \ln (x^2 + 9) - (x^2 + 9) \right) \right]_1^2 \\ &= \frac 12 \left( 13 \ln 13 - 13 \right) - \frac 12 \left( 10 \ln 10 - 10 \right) \\ &= \frac {13}2 \ln 13 - 5 \ln 10 - \frac 32.\end{aligned}$

Problem 13. Evaluate $\displaystyle \int \tan^3 x\, \mathrm dx$ .

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Solution. We solved this before, and solve it again for revision. Using the Pythagorean identity $\sec^2 x = \tan^2 x +1$ ,

$\displaystyle \begin{aligned} \int \tan^3 x\, \mathrm dx &= \int \tan x \tan^2 x\, \mathrm dx \\ &= \int \tan x (\sec^2 x - 1)\, \mathrm dx \\ &= \int \tan x \sec^2 x \, \mathrm dx - \int \tan x\, \mathrm dx \\ &= \int \tan x \sec^2 x \, \mathrm dx -(- \ln|{\cos x}|) + C.\end{aligned}$

For the first integral, make the substitution $\displaystyle u = \tan x \Rightarrow \mathrm dx = \frac{1}{\sec^2 x}\, \mathrm du$ . Then

$\displaystyle \begin{aligned}\int \tan x \sec^2 x \, \mathrm dx &= \int u \sec^2 x \cdot \frac{1}{\sec^2 x}\, \mathrm du \\ &= \int u\, \mathrm du \\ &= \frac 12 u^2 + C\\ &= \frac 12 \tan^2 x + C.\end{aligned}$

Therefore, $\displaystyle \int \tan^3 x\, \mathrm dx = \frac 12 \tan^2 x + \ln|{\cos x}| + C.$

Problem 14. Evaluate $\displaystyle \int \ln (x^2+9)\, \mathrm dx$ .

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Solution. Writing $\ln (x^2+9) = 1 \cdot \ln (x^2+9)$ and integrating by parts,

$\displaystyle \begin{aligned} \int \ln (x^2+9)\, \mathrm dx &= \underbrace{x}_{\mathrm I} \cdot \underbrace{\ln (x^2+9)}_{\mathrm S} - \int \underbrace{x}_{\mathrm I} \cdot \underbrace{\frac{2x}{x^2+9}}_{\mathrm D}\, \mathrm dx \\ &= x \ln(x^2 + 9) - \int \frac{2x^2}{x^2 + 9}\, \mathrm dx \\ &= x \ln(x^2 + 9) - 2\int \frac{(x^2 + 9) - 9}{x^2 + 1}\, \mathrm dx \\ &= x \ln(x^2 + 9) - 2\int \left( 1 - 9 \cdot \frac{1}{3^2 + x^2}\right) \, \mathrm dx \\ &= x \ln(x^2 + 9) - 2 \left( x - 9 \cdot \frac 1{3} \tan^{-1} \left( \frac x3 \right) \right) + C\\ &= x \ln(x^2 + 9) - 2x + 6\tan^{-1} \left( \frac x3 \right) + C.\end{aligned}$

Problem 15. Evaluate $\displaystyle \int \frac {2x+5}{x(x^2+9)}\, \mathrm dx$ .

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Solution. If we can decompose the integrand into

$\displaystyle \frac {2x+5}{x(x^2+9)} = \frac Ax + \frac{Bx+C}{x^2+9},$

Then a bit of basic integration yields

$\displaystyle \begin{aligned} \int \frac {2x+5}{x(x^2+9)}\, \mathrm dx &= \int \left(\frac Ax + \frac{Bx+C}{x^2+9}\right)\, \mathrm dx \\ &= \int \left(A \cdot \frac 1x + \frac B2 \cdot \frac{2x}{x^2+9} + C \cdot \frac 1{x^2 + 9} \right)\, \mathrm dx \\ &= A \int \frac 1x\, \mathrm dx + \frac B2 \int \frac {2x}{x^2 + 1}\, \mathrm dx + C \int \frac 1{3^2 + x^2}\, \mathrm dx \\ &= A \ln|x| + \frac B2 \ln (x^2 + 1) + C \cdot \frac 13 \tan^{-1} \left(\frac x3 \right) + D. \end{aligned}$

It remains to complete the decomposition. Clearing denominators,

$\begin{aligned}\frac {2x+5}{x(x^2+9)} &= \frac Ax + \frac{Bx+C}{x^2+9} \\ 2x+5 &= \frac {Ax(x^2+9)}x + \frac{(Bx+C)x(x^2+9)}{x^2+9} \\ 2x+5 &= A(x^2 + 9) + (Bx + C)x\\ 2x+5 &= (A+B)x^2 + Cx + 9A.\end{aligned}$

Setting $x = 0$ yields $A = 5/9$ .

Comparing coefficients yields $B = -5/9$ , and $C = 2$ .

Hence, by back-substituting,

$\displaystyle \begin{aligned} \int \frac 1{x(x^2+9)}\, \mathrm dx &= \frac 59 \ln|x| - \frac 5{18} \ln (x^2 + 9) + \frac 23 \tan^{-1} \left(\frac x3 \right) + C. \end{aligned}$

Problem 16. Evaluate $\displaystyle \int \frac {(2 \sin \theta +5)\cos \theta}{\sin \theta (\sin^2 \theta+9)}\, \mathrm d\theta$ .

(Click for Solution)

Solution. Making the substitution $\displaystyle u = \sin \theta \Rightarrow \mathrm d\theta = \frac{1}{\cos \theta}\, \mathrm du$ and applying the result from Problem 15,

$\displaystyle \begin{aligned} & \int \frac {(2 \sin \theta +5)\cos \theta}{\sin \theta (\sin^2 \theta+9)}\, \mathrm dx \\ &= \int \frac {(2u+5)\cos \theta}{u (u^2+9)} \cdot \frac{1}{\cos \theta}\, \mathrm du \\ &= \int \frac {2u+5}{u (u^2+9)} \, \mathrm du \\ &= \frac 59 \ln|u| - \frac 5{18} \ln (u^2 + 9) + \frac 23 \tan^{-1} \left(\frac u3 \right) + C \\ &= \frac 59 \ln| {\sin \theta} | - \frac 5{18} \ln ( \sin^2 \theta + 9 ) + \frac 23 \tan^{-1} \left(\frac {\sin \theta}3 \right) + C. \end{aligned}$

—Joel Kindiak, 13 Feb 25, 1757H

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