Calculus — Daily Practice

Monday, July 20, 2026 · 3 problems · Cute Calculus

1.Calculus · Derivatives
Find the derivative of $f(x) = 3x^4 - 5x^2 + 7x - 2$ .
2.Calculus · Indefinite integrals
Evaluate the indefinite integral: $\int \left(6x^2 e^{x^3} + \frac{4}{x}\right)\,dx.$
3.Calculus · Definite integrals
Evaluate the definite integral $\int_{1}^{4} \left(3x^2 - \frac{2}{\sqrt{x}}\right)\, dx.$

Solutions

Calculus · Monday, July 20, 2026

1.Calculus · Derivatives
Find the derivative of $f(x) = 3x^4 - 5x^2 + 7x - 2$ .
1. Apply the power rule to each term: $\frac{d}{dx}[x^n] = n x^{n-1}$ .
2. Differentiate $3x^4$ : $12x^3$ .
3. Differentiate $-5x^2$ : $-10x$ .
4. Differentiate $7x$ : $7$ .
5. Differentiate the constant $-2$ : $0$ .
6. Combine the results: $f'(x) = 12x^3 - 10x + 7$ .
Answer:
$f'(x) = 12x^3 - 10x + 7$
2.Calculus · Indefinite integrals
Evaluate the indefinite integral: $\int \left(6x^2 e^{x^3} + \frac{4}{x}\right)\,dx.$
1. Split the integral into two parts: $\int 6x^2 e^{x^3}\,dx + \int \frac{4}{x}\,dx$ .
2. For the first integral, use the substitution $u = x^3$ , so $du = 3x^2\,dx$ , which gives $6x^2\,dx = 2\,du$ .
3. Thus $\int 6x^2 e^{x^3}\,dx = \int 2 e^u\,du = 2e^u = 2e^{x^3}$ .
4. For the second integral, $\int \frac{4}{x}\,dx = 4\ln|x|$ .
5. Combine the results and add the constant of integration: $2e^{x^3} + 4\ln|x| + C$ .
Answer:
$2e^{x^3} + 4\ln|x| + C$
3.Calculus · Definite integrals
Evaluate the definite integral $\int_{1}^{4} \left(3x^2 - \frac{2}{\sqrt{x}}\right)\, dx.$
1. Split the integral: $\int_{1}^{4} 3x^2\,dx - \int_{1}^{4} 2x^{-1/2}\,dx$ .
2. Antiderivative of $3x^2$ is $x^3$ .
3. Antiderivative of $2x^{-1/2}$ is $2 \cdot \frac{x^{1/2}}{1/2} = 4x^{1/2} = 4\sqrt{x}$ .
4. So the antiderivative is $F(x) = x^3 - 4\sqrt{x}$ .
5. Evaluate at upper bound: $F(4) = 4^3 - 4\sqrt{4} = 64 - 8 = 56$ .
6. Evaluate at lower bound: $F(1) = 1^3 - 4\sqrt{1} = 1 - 4 = -3$ .
7. Subtract: $56 - (-3) = 59$ .
Answer:
$59$