In this article, we will review the final exam of Calculus 1 in detail and chapterwise. The syllabus of Calculus 1 includes Limit, Continuity, Derivative, Integration, and their applications.
Table of Contents
Limit Final Exam Review
Question 1: Find $\lim\limits_{x \to 2} \dfrac{x^2-5x+6}{x^2-4}$.
Solution:
At first, we will factorize both the numerator and the denominator. Note that x2-5x+6 = x2-3x-2x+6 = x(x-3)-2(x-3) = (x-2)(x-3).
And x2-4 = (x-2)(x+2) by the formula a2-b2 = (a-b)(a+b).
$\therefore \dfrac{x^2-5x+6}{x^2-4}$ $=\dfrac{(x-2)(x-3)}{(x-2)(x+2)}$ $=\dfrac{x-3}{x+2}$
Thus, the limit will be equal to
$\lim\limits_{x \to 2} \dfrac{x-3}{x+2}$
= $\dfrac{2-3}{2+2}=-\dfrac{1}{4}$.
Limit: Definition, Formulas, Examples
Continuity Final Exam Review
Question 2: Find the value of c if the following function f(x) is continuous at x=1.
$f(x)=\begin{cases}cx-3 \quad \quad \text{ if } x\geq 1 \\ x^2+2cx \quad \quad \text{ if } x<1 \end{cases}$
Solution:
As f(x) is continuous at x=1, we must have that
limx→1+ f(x) = limx→1- f(x) = f(1).
Therefore, [cx-3]x=1 = [x2+2cx]x=1
⇒ c⋅1-3 = 12+2c⋅1
⇒ c-3 = 1+2c
⇒ 2c-c = -3-1 ⇒ c=-4.
Thus, if the value of c is -4 then the function f(x) will be continuous at x=1.
Must Read:
Continuity of a Function: Definition, Properties, Solved Examples
Discontinuity of a Function: Definition, Types, Examples
Derivative Final Exam Review
Question 3: Find the derivative of f(x) = x(x-1)(x-2).
Solution:
We have x(x-1)(x-2) = x(x2-x-2x+2) = x(x2-3x+2) = x3-3x2+2x.
So $\dfrac{d}{dx}\{x(x-1)(x-2)\}$
= $\dfrac{d}{dx}(x^3-3x^2+2x)$
= $\dfrac{d}{dx}(x^3)-3\dfrac{d}{dx}(x^2)+2\dfrac{d}{dx}(x)$
= 3x2-6x+2 by the power rule of derivative
Also Read:
Derivative: Definition, Formulas, Examples
Integration Final Exam Review
Question 4: Find the integral ∫ (√x+x3)dx
Solution:
We have ∫ (√x+x3)dx
= ∫x1/2dx + ∫x3dx
= $\dfrac{x^{1/2+1}}{\frac{1}{2}+1}+\dfrac{x^4}{4}+C$
= 2/3 x3/2 +x4/4+C, where C is an integral constant.
Also Read:
