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 x^{2}-5x+6 = x^{2}-3x-2x+6 = x(x-3)-2(x-3) = (x-2)(x-3).

And x^{2}-4 = (x-2)(x+2) by the formula a^{2}-b^{2} = (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

lim_{x→1+} f(x) = lim_{x→1- }f(x) = f(1).

Therefore, [cx-3]_{x=1} = [x^{2}+2cx]_{x=1}

⇒ c⋅1-3 = 1^{2}+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(x^{2}-x-2x+2) = x(x^{2}-3x+2) = x^{3}-3x^{2}+2x.

So $\dfrac{d}{dx}\{x(x-1)(x-2)\}$

= $\dfrac{d}{dx}(x^3-3x^2+2x)$

= $\dfrac{d}{dx}(x^3)-3\dfrac{d}{dx}(x^2)+2\dfrac{d}{dx}(x)$

= 3x^{2}-6x+2 by the power rule of derivative

**Also Read:**

**Derivative: Definition, Formulas, Examples**

## Integration Final Exam Review

**Question 4:** Find the integral ∫ (√x+x^{3})dx

**Solution:**

We have ∫ (√x+x^{3})dx

= ∫x^{1/2}dx + ∫x^{3}dx

= $\dfrac{x^{1/2+1}}{\frac{1}{2}+1}+\dfrac{x^4}{4}+C$

= 2/3 x^{3/2} +x^{4}/4+C, where C is an integral constant.

**Also Read:**