Question:
\(\int^1_0\frac{1}{\sqrt{3+x}+\sqrt{1+x}}dx=a+b\sqrt2+c\sqrt3\) then \(2a-3b-4c\) is equal to _____.
\(\int^1_0\frac{1}{\sqrt{3+x}+\sqrt{1+x}}dx=a+b\sqrt2+c\sqrt3\) then \(2a-3b-4c\) is equal to _____.
Updated On: May 14, 2024
- 10
- 0
- 12
- 20
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Hence, The correct option is (C): \(12\)
Was this answer helpful?
2
13
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 $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
- JEE Main - 2025
- Relations and functions
- If \( \alpha + i\beta \) and \( \gamma + i\delta \) are the roots of the equation \( x^2 - (3-2i)x - (2i-2) = 0 \), \( i = \sqrt{-1} \), then \( \alpha\gamma + \beta\delta \) is equal to:
- JEE Main - 2025
- Complex numbers
- The sum of all local minimum values of the function \( f(x) \) as defined below is:
\[ f(x) = \begin{cases} 1 - 2x & \text{if } x < -1, \\[10pt] \frac{1}{3}(7 + 2|x|) & \text{if } -1 \leq x \leq 2, \\[10pt] \frac{11}{18}(x-4)(x-5) & \text{if } x > 2. \end{cases} \]- JEE Main - 2025
- Functions
- Let \( \langle a_n \rangle \) be a sequence such that \( a_0 = 0 \), \( a_1 = \frac{1}{2} \), and \( 2a_{n+2} = 5a_{n+1} - 3a_n \).n= 0,1,2,3.... Then \( \sum_{k=1}^{100} a_k \) is equal to:
- JEE Main - 2025
- Sequences and Series
- The expression given below shows the variation of velocity \( v \) with time \( t \): \[ v = At^2 + \frac{Bt}{C + t}. \] The dimension of \( ABC \) is:
- JEE Main - 2025
- Thermodynamics terms
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.