# Calculus 1 Final Exam Review

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.

## 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.

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

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.