Top 10 Calculus Problems Every Student Should Know (with Solutions)

Calculus can be a challenging but rewarding subject. The best way to master its concepts is through consistent practice. This guide presents 10 fundamental calculus problems, ranging from limits to integrals, that every student should be able to solve. Each problem comes with a detailed, step-by-step solution to help solidify your understanding.

1. Limits

Problem 1: Evaluate the Limit (Algebraic Manipulation)

Evaluate the limit:

$$\lim_{x \to 4} \frac{x^2 - 16}{x - 4}$$

Solution: If we try direct substitution, we get $\frac{4^2 - 16}{4 - 4} = \frac{0}{0}$, which is an indeterminate form. This means we need to manipulate the expression.

1. Factor the numerator: The numerator is a difference of squares. $$x^2 - 16 = (x - 4)(x + 4)$$
2. Substitute the factored form back into the limit: $$\lim_{x \to 4} \frac{(x - 4)(x + 4)}{x - 4}$$
3. Cancel out the common factor: Since $x \to 4$, $x \neq 4$, so $x - 4 \neq 0$. We can cancel $(x - 4)$. $$\lim_{x \to 4} (x + 4)$$
4. Evaluate the limit by direct substitution: $$4 + 4 = 8$$

Answer: $8$

2. Derivatives (Power Rule & Chain Rule)

Problem 2: Find the Derivative (Combined Rules)

Find the derivative of $f(x) = (3x^2 - 5x + 1)^4$.

Solution: This function is a composite function, so we need to use the Chain Rule, combined with the Power Rule.

1. Identify the outer function and inner function:
   • Outer function: $u^4$
   • Inner function: $u = 3x^2 - 5x + 1$
2. Apply the Chain Rule: $\frac{d}{dx} [g(u)] = g'(u) \cdot u'$
   • Derivative of the outer function: $4u^3$
   • Derivative of the inner function: $\frac{d}{dx}(3x^2 - 5x + 1) = 6x - 5$
3. Substitute back: $$f'(x) = 4(3x^2 - 5x + 1)^3 (6x - 5)$$

Answer: $f'(x) = 4(6x - 5)(3x^2 - 5x + 1)^3$

3. Derivatives (Product Rule)

Problem 3: Find the Derivative (Product Rule)

Find the derivative of $g(x) = x^3 \sin x$.

Solution: This is a product of two functions ($u = x^3$ and $v = \sin x$), so we use the Product Rule: $\frac{d}{dx}(uv) = u'v + uv'.$

1. Identify $u$ and $v$ and their derivatives:
   • $u = x^3 \implies u' = 3x^2$
   • $v = \sin x \implies v' = \cos x$
2. Apply the Product Rule: $$g'(x) = (3x^2)(\sin x) + (x^3)(\cos x)$$

Answer: $g'(x) = 3x^2 \sin x + x^3 \cos x$

4. Derivatives (Quotient Rule)

Problem 4: Find the Derivative (Quotient Rule)

Find the derivative of $h(x) = \frac{e^x}{x^2 + 1}$.

Solution: This is a quotient of two functions, so we use the Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$.

1. Identify $u$ and $v$ and their derivatives:
   • $u = e^x \implies u' = e^x$
   • $v = x^2 + 1 \implies v' = 2x$
2. Apply the Quotient Rule: $$h'(x) = \frac{(e^x)(x^2 + 1) - (e^x)(2x)}{(x^2 + 1)^2}$$
3. Simplify the numerator: $$h'(x) = \frac{e^x(x^2 + 1 - 2x)}{(x^2 + 1)^2} = \frac{e^x(x - 1)^2}{(x^2 + 1)^2}$$

Answer: $h'(x) = \frac{e^x(x - 1)^2}{(x^2 + 1)^2}$

5. Implicit Differentiation

Problem 5: Implicit Differentiation

Find $\frac{dy}{dx}$ for the equation $x^2 + y^2 = 25$.

Solution: Since $y$ is implicitly defined as a function of $x$, we differentiate both sides of the equation with respect to $x$, remembering to apply the Chain Rule to terms involving $y$.

1. Differentiate each term with respect to $x$:
   • $\frac{d}{dx}(x^2) = 2x$
   • $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$ (using the Chain Rule)
   • $\frac{d}{dx}(25) = 0$
2. Substitute these back into the equation: $$2x + 2y \frac{dy}{dx} = 0$$
3. Isolate $\frac{dy}{dx}$: $$2y \frac{dy}{dx} = -2x$$ $$\frac{dy}{dx} = \frac{-2x}{2y} = -\frac{x}{y}$$

Answer: $\frac{dy}{dx} = -\frac{x}{y}$

6. Related Rates

Problem 6: Related Rates (Expanding Circle)

The radius of a circle is increasing at a rate of $2 \, \text{cm/s}$. How fast is the area of the circle increasing when the radius is $10 \, \text{cm}$?

Solution:

1. Identify given information and what needs to be found:
   • Given: $\frac{dr}{dt} = 2 \, \text{cm/s}$
   • Find: $\frac{dA}{dt}$ when $r = 10 \, \text{cm}$
2. Write down the relevant formula: The area of a circle is $A = \pi r^2$.
3. Differentiate the formula with respect to time ($t$): $$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt}$$
4. Substitute the known values: $$\frac{dA}{dt} = 2\pi (10 \, \text{cm}) (2 \, \text{cm/s})$$ $$\frac{dA}{dt} = 40\pi \, \text{cm}^2/\text{s}$$

Answer: The area of the circle is increasing at a rate of $40\pi \, \text{cm}^2/\text{s}$.

7. Optimization

Problem 7: Optimization (Maximum Area)

A farmer has $1200 \, \text{ft}$ of fencing and wants to fence off a rectangular field bordering a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area?

Solution:

1. Define variables and set up equations:
   • Let $l$ be the length of the field parallel to the river and $w$ be the width perpendicular to the river.
   • Perimeter: $l + 2w = 1200$ (no fence along the river)
   • Area: $A = lw$
2. Express Area as a function of one variable: From the perimeter equation, $l = 1200 - 2w$. Substitute this into the area equation. $$A(w) = (1200 - 2w)w = 1200w - 2w^2$$
3. Find the critical points by taking the derivative and setting it to zero: $$A'(w) = 1200 - 4w$$ Set $A'(w) = 0$: $$1200 - 4w = 0 \implies 4w = 1200 \implies w = 300 \, \text{ft}$$
4. Determine the corresponding length: $$l = 1200 - 2(300) = 1200 - 600 = 600 \, \text{ft}$$
5. Verify it's a maximum (using the second derivative test): $$A''(w) = -4$$ Since $A''(w) < 0$, the critical point corresponds to a local maximum.

Answer: The dimensions that maximize the area are $600 \, \text{ft}$ (length parallel to the river) by $300 \, \text{ft}$ (width perpendicular to the river).

8. Indefinite Integral (Basic)

Problem 8: Basic Indefinite Integral

Evaluate $\int (4x^3 - 6x^2 + \cos x) \, dx$.

Solution: We can integrate each term separately using the power rule for integration and standard trigonometric integrals.

1. Integrate each term:
   • $\int 4x^3 \, dx = 4 \frac{x^{3+1}}{3+1} = 4 \frac{x^4}{4} = x^4$
   • $\int -6x^2 \, dx = -6 \frac{x^{2+1}}{2+1} = -6 \frac{x^3}{3} = -2x^3$
   • $\int \cos x \, dx = \sin x$
2. Combine the results and add the constant of integration $C$: $$\int (4x^3 - 6x^2 + \cos x) \, dx = x^4 - 2x^3 + \sin x + C$$

Answer: $x^4 - 2x^3 + \sin x + C$

9. Definite Integral (Area Under a Curve)

Problem 9: Definite Integral

Evaluate $\int_1^3 (x^2 + 2x) \, dx$.

Solution:

1. Find the indefinite integral (antiderivative): $$\int (x^2 + 2x) \, dx = \frac{x^3}{3} + x^2$$
2. Apply the Fundamental Theorem of Calculus (evaluate at upper and lower limits): $$\left[\frac{x^3}{3} + x^2\right]_1^3 = \left(\frac{3^3}{3} + 3^2\right) - \left(\frac{1^3}{3} + 1^2\right)$$ $$= \left(\frac{27}{3} + 9\right) - \left(\frac{1}{3} + 1\right)$$ $$= (9 + 9) - \left(\frac{1}{3} + \frac{3}{3}\right)$$ $$= 18 - \frac{4}{3}$$
3. Combine terms: $$= \frac{54}{3} - \frac{4}{3} = \frac{50}{3}$$

Answer: $\frac{50}{3}$

10. Substitution (u-Substitution)

Problem 10: U-Substitution

Evaluate $\int x \sqrt{x^2 + 5} \, dx$.

Solution: This integral is suitable for u-substitution because we have a function ($x^2 + 5$) and its derivative (or a multiple of its derivative, $x$) present in the integrand.

1. Choose $u$ and find $du$:
   • Let $u = x^2 + 5$
   • Then $du = 2x \, dx \implies \frac{1}{2} du = x \, dx$
2. Substitute $u$ and $du$ into the integral: $$\int \sqrt{u} \left(\frac{1}{2} du\right) = \frac{1}{2} \int u^{1/2} \, du$$
3. Integrate with respect to $u$: $$= \frac{1}{2} \left(\frac{u^{1/2 + 1}}{1/2 + 1}\right) + C$$ $$= \frac{1}{2} \left(\frac{u^{3/2}}{3/2}\right) + C$$ $$= \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3} u^{3/2} + C$$
4. Substitute back $x^2 + 5$ for $u$: $$= \frac{1}{3} (x^2 + 5)^{3/2} + C$$

Answer: $\frac{1}{3} (x^2 + 5)^{3/2} + C$

Conclusion

Working through these types of problems will significantly strengthen your calculus skills. Remember to understand the underlying concepts, not just memorize the steps. Consistent practice and reviewing solutions are key to success in calculus!