Question:
The value of integrate \(∫^1_0 (2x^3 - 3x^2 - x + 1)^\frac{1}{3}dx\) is :
The value of integrate \(∫^1_0 (2x^3 - 3x^2 - x + 1)^\frac{1}{3}dx\) is :
Updated On: Mar 20, 2025
- -1
- 1
- 0
- 2
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
The Correct Option is (C): 0
Was this answer helpful?
2
11
Top Questions on integral
- Evaluate the integral: $$ \int_0^{\pi/4} \frac{\ln(1 + \tan x)}{\cos x \sin x} \, dx $$
- Find the value of the integral: \[ \int_0^\pi \sin^2(x) \, dx. \]
- Evaluate the integral \( \int_0^1 \frac{\ln(1 + x)}{1 + x^2} \, dx \)
- Evaluate the integral: $$ \int_0^1 \frac{\ln (1+x)}{1+x^2} \, dx $$
- Evaluate $ \int_{0}^{1} (3x^2 + 2x) \, dx $.
View More Questions
Questions Asked in JEE Main exam
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to:
- JEE Main - 2025
- Matrix Operations
- Let \( P_n = \alpha^n + \beta^n \), \( P_{10} = 123 \), \( P_9 = 76 \), \( P_8 = 47 \), and \( P_1 = 1 \). The quadratic equation whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is:
- JEE Main - 2025
- Sequences and Series
- Number of functions \( f: \{1, 2, \dots, 100\} \to \{0, 1\} \), that assign 1 to exactly one of the positive integers less than or equal to 98, is equal to:
- JEE Main - 2025
- Vectors
- Three infinitely long wires with linear charge density \( \lambda \) are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?
- JEE Main - 2025
- electrostatic potential and capacitance
- The number of 6-letter words, with or without meaning, that can be formed using the letters of the word "MATHS" such that any letter that appears in the word must appear at least twice is:
- JEE Main - 2025
- Calculus
View More Questions
Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.