Basic Mathematics · Unit 6
Integration and Antiderivatives
Exam-focused notes for Integration and Antiderivatives (Basic Mathematics, MTH104): what the TU syllabus asks and how it has actually been tested, with 8 solved past questions from this unit.
What this unit covers
- Indefinite integral definition
- Basic integration rules
- Integration by substitution
- Integration by parts
- Partial fraction decomposition
- Trigonometric integrals
- Definite integral definition
- Fundamental theorem of calculus
- Improper integrals
Trigonometric integrals
Evaluate: $$\int \sqrt{4 - x^2} , dx$$ [5]
- Integral to evaluate: $\displaystyle \int \sqrt{4 - x^2}\, dx$ - Form: $\sqrt{a^2 - x^2}$ with $a^2 = 4 \Rightarrow a = 2$ Let $x = 2\sin\theta \Rightarrow dx = 2\cos\theta\, d\theta$. Then: $$\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = \sqrt{4\cos^2\thet...
Full solved answer →Evaluate $\int_0^{\frac{\pi}{2}}(\sin 2x\cos 3x + \cos 2x\sin 3x)dx$ and $\int_{0}^{\frac{\pi}{4}} \frac{dx}{1 - \sin x}$. [5]
- Integral 1: $\displaystyle\int0^{\pi/2}(\sin 2x\cos 3x + \cos 2x\sin 3x)\,dx$ - Integral 2: $\displaystyle\int0^{\pi/4}\frac{dx}{1-\sin x}$ --- Step 1: Apply sine addition formula $$\sin A\cos B + \cos A\sin B = \sin(A+B)$$ With $A = 2x$, $B = 3x$: $$\sin...
Full solved answer →Integrate the following: $$\int_{0}^{\frac{\pi}{4}} \frac{dx}{1-\sin x}$$ [5]
- Integrand: $\dfrac{1}{1-\sin x}$ - Limits: from $0$ to $\dfrac{\pi}{4}$ (the LaTeX "x/4" is a rendering artifact for the standard upper limit $\pi/4$) $$I = \int{0}^{\pi/4} \frac{dx}{1-\sin x}$$ Multiply numerator and denominator by the conjugate $(1+\sin...
Full solved answer →Improper integrals
Evaluate: $$\int_0^5 \frac{dx}{\sqrt{x - 2}}$$, if it exists. [5]
- Integrand: $f(x) = \dfrac{1}{\sqrt{x-2}}$ - Lower limit: $0$ - Upper limit: $5$ The integrand $\dfrac{1}{\sqrt{x-2}}$ is only defined (in real numbers) when $x - 2 0$, i.e. $x 2$. For $x < 2$, the quantity $x - 2 < 0$, so $\sqrt{x-2}$ is not a real number...
Full solved answer →Integration by parts
Evaluate: $$\int_0^{\pi} x \sin x , dx$$ [5]
- Integrand: $x \sin x$ - Limits: from $0$ to $\pi$ Using $\int u \, dv = uv - \int v \, du$. Let: - $u = x \Rightarrow du = dx$ - $dv = \sin x \, dx \Rightarrow v = -\cos x$ $$\int0^{\pi} x \sin x \, dx = \Big[-x\cos x\Big]0^{\pi} - \int0^{\pi} (-\cos x)\,...
Full solved answer →Evaluate $\int_{0}^{\frac{\pi}{4}} \frac{dx}{1-\sin x}$, Evaluate $\int x^2 \sin x , dx$. Solve the differential equation $\frac{dy}{dx} - \frac{3y}{x} = x$, $x > 0$. [5+5]
Given data: Integrand $\frac{1}{1-\sin x}$, limits $0$ to $\pi/4$. Multiply numerator and denominator by the conjugate $(1+\sin x)$: $$\int0^{\pi/4} \frac{1+\sin x}{(1-\sin x)(1+\sin x)}\,dx = \int0^{\pi/4} \frac{1+\sin x}{1-\sin^2 x}\,dx = \int0^{\pi/4} \f...
Full solved answer →Basic integration rules
Evaluate the following integral. $$\int_{0}^{\frac{\pi}{4}} \sqrt{1 + \cos x}dx$$ [5]
- Integral 1: $\displaystyle\int{0}^{\pi/4} \sqrt{1+\cos x}\,dx$ - Integral 2: $\displaystyle\int \frac{3x^2-7x+1}{3x}\,dx$ --- Step 1: Simplify using half-angle identity $$1 + \cos x = 2\cos^2\frac{x}{2}$$ So: $$\sqrt{1+\cos x} = \sqrt{2}\left\cos\frac{x}{...
Full solved answer →Partial fraction decomposition
Define integration. Evaluate the following integral. $\int \frac{dx}{(x-1)(x-2)}$, $\int_{-1}^{1} 3x^2 \sqrt{x^3 + 1} ,dx$ [1+4]
Integration is the reverse process of differentiation. It is the process of finding a function (the antiderivative) whose derivative is a given function. If $F'(x) = f(x)$, then: $$\int f(x)\,dx = F(x) + C$$ where $C$ is the constant of integration. For a d...
Full solved answer →Make Unit 6 stick
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