Calculus is a branch of mathematics that deals with the study of rates of change and accumulation of quantities. It provides a framework for modeling and analyzing problems involving change, such as motion, growth, and decay.

There are two main branches of calculus:

• Differential Calculus: This branch focuses on the concept of the derivative, which represents the rate of change of a function at a particular point.
• Integral Calculus: Integral calculus deals with the concept of the integral, which represents the accumulation of quantities and the computation of areas under curves.

Solving Practice Questions on Calculus is the best way to master various concepts of calculus easily.

This article contains various Calculus Practice Questions on various topics including Derivatives, Integrals, Application of Derivatives and Mean Value Theorem.

Limit of a Function

In mathematics, for a given function f(x), where x is a real number, we say that L is the limit of the function of at x = a if, whenever x approaches a, f(x) approaches L. This is written as:

lim_⁡x→af(x)=L

To understand this concept, let’s look at an example.

Example: Find lim_x→2 f(x), where f(x) = (x² − 4)/(x − 2).

Consider the function f(x) = (x² − 4)/(x − 2)​.

To find lim⁡_x→2f(x), where f(x) = (x² − 4)/(x − 2)​, we can simplify the expression first:

f(x) = (x² − 4)/(x − 2)

f(x) = (x − 2)(x + 2)/(x − 2)

f(x) = (x + 2)

Now, the limit becomes:

lim⁡_x→2f(x) = (x + 2)

Plugging in x = 2, we get:

lim⁡_x→2f(x) = 2 + 2 =4

So, lim⁡_x→2f(x) = 4.

This means that even though f(x) is not defined at x = 2, we can determine that the function approaches the value 4 as x gets closer and closer to 2.

Continuity of a Function

A function f(x) is said to be continuous at a point (x = a) if three conditions are satisfied:

• f(a) is defined.
• lim⁡_x→a f(x) exists.
• lim⁡_x→a f(x) = f(a)).

In simpler terms, a function is continuous at a point if there is no interruption in its graph at that point. The function must not have any holes, jumps, or vertical asymptotes at that point.

Examples of Continuity

• Polynomials: Functions like ( f(x) = x² + 3x + 2 ) are continuous everywhere because they are polynomials.
• Rational Functions: Functions like ( f(x) = (x² – 1)/(x – 1) are continuous everywhere except where the denominator is zero.

Calculus Practice Questions – Derivatives

1: Find the derivative of the function f(x) = 3x² + 4x – 2.

We can find the derivative of f(x) using the power rule. The power rule states that if f(x) = axⁿ, then f'(x) = nax^(n-1).

Applying the power rule:

f'(x) = 2 x 3x^(2-1)+ 1 x4x^(1-1) – 0 x 2 = 6x + 4.

2: Find the derivative of the function g(x) = 1/x².

We can use the power rule and the chain rule for this problem. Rewrite g(x) as g(x) = x^(-2).

Applying the power rule and chain rule:

g'(x) = -2x^(-3) = -2/x³.

3: Find the derivative of the function f(x) = 5x³ – 2x² + 7x – 9.

Using the power rule:

f'(x) = 3 x 5x^(3-1) – 2 × 2x^(2-1) + 7 × 1x^(1-1)– 0 = 15x² – 4x + 7.

4: Find the derivative of the function g(x) = 2/x³ – 4x² + (x).

Using the power rule and the constant multiple rule:

g'(x) = -3 × 2/x³⁺¹ – 4 × 2x^2-1 + 1/(2 × √(x)) = -6/x⁴ – 8x + 1/(2 × √(x)).

5: Find the derivative of the function h(x) = e^2x + ln(x).

The derivative of e^2x is 2e^2x by the chain rule, and the derivative of ln(x) is 1/x.

So, h'(x) = 2e^2x+ 1/x.

6: Find the derivative of the function f(x) = sin(x) + cos(x).

The derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x).

So, f'(x) = cos(x) – sin(x).

7: Find the derivative of the function g(x) = x² × e^x.

Apply the product rule:

g'(x) = (2x × e^x) + (x² × e^x) = x² × e^x + 2x × e^x.

8: Find the derivative of the function h(x) = (x² + 1) / (x³ + x).

Apply the quotient rule:

h'(x) = [(2x × (x³ + x)) – ((x² + 1) × (3x² + 1))] / (x³ + x)².

9: Find the derivative of f(x) = (3x² + 2x – 1)/(x² – 4).

Using the quotient rule for differentiation:

Let u(x) = 3x² + 2x – 1 and v(x) = x² – 4

Then, f(x) = u(x)/v(x)

f'(x) = (u'(x)v(x) – u(x)v'(x)) / (v(x))²

u'(x) = 6x + 2

v'(x) = 2x

Substituting and simplifying:

f'(x) = ((6x + 2)(x² – 4) – (3x² + 2x – 1)(2x)) / (x² – 4)²

⇒ f'(x) = (6x³ + 2x² – 24x – 8 – 6x³ – 4x² + 2x) / (x² – 4)²

⇒ f'(x) = (-4x² – 22x – 8) / (x² – 4)²

Calculus Practice Questions – Integrals

1: Evaluate the definite integral: ∫(2 to 5) (x³ – 2x² + x + 3) dx

To evaluate a definite integral, we need to find the antiderivative (indefinite integral) first, and then evaluate it at the given limits.

Let F(x) = ∫(x³ – 2x² + x + 3) dx

F(x) = (x⁴/4) – (2x³/3) + (x²/2) + 3x + C

Substituting the limits:

∫(2 to 5) (x³ – 2x² + x + 3) dx = F(5) – F(2)

= [(5⁴/4) – (2(5³)/3) + (5²/2) + 3(5)] – [(2⁴/4) – (2(2³)/3) + (2²/2) + 3(2)]

= (625/4 – 250/3 + 25/2 + 15) – (16/4 – 16/3 + 2 + 6)

= 156.25 – 83.33 + 12.5 + 15 – 4 – 5.33 + 2 + 6

= 99.17

2: Evaluate the indefinite integral: ∫(x² + 3x + 2)/(x – 1)³ dx

To evaluate an indefinite integral, we need to find the antiderivative or the integral function.

Let u = x – 1, then du = dx

Substituting u + 1 for x, we get:

∫(u² + 3u + 3)/(u)³

3: Evaluate the definite integral: ∫(1 to e) (ln(x))² dx

To evaluate a definite integral, we need to find the antiderivative (indefinite integral) first, and then evaluate it at the given limits.

Let F(x) = ∫(ln(x))² dx

Using the substitution u = ln(x), du = (1/x) dx

F(x) = ∫u² du

⇒ F(x) = u³/3 + C

⇒ F(x) = (ln(x))³/3 + C

Substituting the limits:

∫(1 to e) (ln(x))² dx = F(e) – F(1)

= (ln(e))^3/3 – (ln(1))^3/3

= 1/3 – 0

= 1/3

Calculus Practice Questions – Application of Derivatives

1: Find the equation of the tangent line to the curve y = x³ – 3x² + 2x + 1 at the point (1, -1).

To find the equation of the tangent line, we need to calculate the derivative of the function and evaluate it at the given point to find the slope of the tangent line.

y’ = 3x² – 6x + 2

At x = 1, y’ = 3(1)² – 6(1) + 2 = -1

The slope of the tangent line is -1.

The equation of a line is y – y₀ = m(x – x₀), where m is the slope, and (x₀, y₀) is a point on the line.

Substituting the values, we get:

y – (-1) = -1(x – 1)

⇒ y + 1 = -x + 1

⇒ y = -x

2: Find the absolute maximum and minimum values of the function f(x) = x³ – 3x² + 2x + 4 on the closed interval [-2, 3].

To find the absolute maximum and minimum values, we need to:

1. Find the critical points of the function on the given interval.
2. Evaluate the function at the critical points and the endpoints of the interval.
3. The absolute maximum and minimum values will be the largest and smallest values from step 2, respectively.

Step 1: Find the critical points

f'(x) = 3x² – 6x + 2 = 0

⇒ 3x² – 6x + 2 = 0

⇒ 3(x² – 2x + 2/3) = 0

⇒ 3(x – 1)(x – 1) = 0

⇒ x = 1 (repeated root)

Step 2: Evaluate the function at the critical point and the endpoints

f(-2) = (-2)³ – 3(-2)² + 2(-2) + 4 = -8 + 12 – 4 + 4 = 4

f(1) = 1³ – 3(1)² + 2(1) + 4 = 1 – 3 + 2 + 4 = 4

f(3) = 3³ – 3(3)² + 2(3) + 4 = 27 – 27 + 6 + 4 = 10

Step 3: Determine the absolute maximum and minimum values

The minimum value is 4 (occurring at x = -2 and x = 1), and the maximum value is 10 (occurring at x = 3).

3: Find the area between the curves y = x^2 and y = 2x on the interval [0, 2].

To find the area between two curves, we need to integrate the difference between the two functions over the given interval.

Let A be the area between the curves.

A = ∫(0 to 2) (2x – x²) dx

⇒ A = [x² – x³/3]_{0 to 2}

⇒ A = (2² – 2³/3) – (0² – 0³/3)

⇒ A = 4 – 8/3 – 0

⇒ A = 8/3

4: Find the volume of the solid obtained by rotating the region bounded by the curves y = x² and y = 4x about the x-axis.

To find the volume of a solid of revolution, we need to integrate the area of the cross-section at each x-value and then multiply by 2π.

The region is bounded by y = x² and y = 4x from x = 0 to x = 2.

Let V be the volume of the solid.

V = 2π ∫(0 to 2) (4x – x²) dx (Area between the curves)

⇒ V = 2π [2x² – x³/3]_{(0 to 2)}

⇒ V = 2π [(2(2)² – (2)³/3) – (2(0)² – 0³/3)]

⇒ V = 2π (8 – 8/3 – 0)

⇒ V = 2π (24/3)

⇒ V = 16π

Therefore, the volume of the solid obtained by rotating the region bounded by y = x² and y = 4x about the x-axis is 16π cubic units.

5: Use the second derivative test to determine the nature of the critical points of the function f(x) = x⁴ – 4x³ + 2x² + 6x – 3.

Step 1: Find the critical points by setting the first derivative equal to zero.

f'(x) = 4x³ – 12x² + 4x + 6 = 0

⇒ 4x(x² – 3x + 3/2) + 6(1 – x) = 0

⇒ 4x(x – 1)(x – 3/2) + 6(1 – x) = 0

Thus, x = 1, x = 3/2

Step 2: Evaluate the second derivative at the critical points.

f”(x) = 12x² – 24x + 4

⇒ f”(1) = 12 – 24 + 4 = -8 (negative)

⇒ f”(3/2) = 18 – 36 + 4 = -14 (negative)

Step 3: Apply the second derivative test.

• If f”(x) < 0, then the point is a local maximum.
• If f”(x) > 0, then the point is a local minimum.

For x = 1, f”(1) < 0, so x = 1 is a local maximum.

For x = 3/2, f”(3/2) < 0, so x = 3/2 is a local maximum.

6: Use the second derivative test to determine the nature of the critical points of the function f(x) = x⁴ – 4x³ + 2x² + 6x – 3.

Step 1: Find the critical points by setting the first derivative equal to zero.

f'(x) = 4x³ – 12x² + 4x + 6 = 0

⇒ 4x(x² – 3x + 3/2) + 6(1 – x) = 0

⇒ 4x(x – 1)(x – 3/2) + 6(1 – x) = 0

Thus, x = 1, x = 3/2

Step 2: Evaluate the second derivative at the critical points.

f”(x) = 12x² – 24x + 4

⇒ f”(1) = 12 – 24 + 4 = -8 (negative)

⇒ f”(3/2) = 18 – 36 + 4 = -14 (negative)

Step 3: Apply the second derivative test.

• If f”(x) < 0, then the point is a local maximum.
• If f”(x) > 0, then the point is a local minimum.

For x = 1, f”(1) < 0, so x = 1 is a local maximum.

For x = 3/2, f”(3/2) < 0, so x = 3/2 is a local maximum.

Calculus Practice Questions – Mean Value Theorem

1: Use the mean value theorem to show that there exists a point c in the interval (0, 1) such that cos(c) = sin(1).

The mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) such that f'(c) = (f(b) – f(a))/(b – a).

Let f(x) = cos(x) – sin(1 – x)

f(0) = 1 – sin(1)

f(1) = cos(1) – sin(0) = cos(1)

f'(x) = -sin(x) + cos(1 – x)

According to the mean value theorem, there exists a point c in (0, 1) such that:

f'(c) = (f(1) – f(0))/(1 – 0)

⇒ -sin(c) + cos(1 – c) = cos(1) – (1 – sin(1))

⇒ cos(1 – c) = sin(1)

Practice Problems on Calculus – Unsolved

1: Find the derivative of the function f(x) = 3x² + 2x + 1.

2: Calculate the limit: lim_{x → 3}(x² – 9)/(x – 3).

3: Find the critical points of the function g(x) = x³ – 12x + 1.

4: Determine the indefinite integral of the function h(x) = 2x³ + 3x² + 5.

5: Find the area bounded by the curve y = x² and the x-axis from x = 0 to x = 2.

6: Calculate the derivative of the function y = 1/x + 1/x².

7: Find the second derivative of the function f(x) = 5x³ – 4x² + 3x – 2.

8: Determine the limit: lim_{x → 3} sin(x)/x.

9: Find the volume of the solid generated by revolving the region bounded by y = x² and the x-axis from x = 0 to x = 1 about the y-axis.

10: Calculate the derivative of the function y = ln x² + √x.

Read More,

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