Let \([t]\) denote the greatest integer less than or equal to \(t\). Then, the value of the integral \(∫_0^1 [ -8x^2 + 6x - 1] dx\) is equal to
\(-1\)
\(-\frac 54\)
\(\frac {\sqrt {17}-13}{8}\)
\(\frac {\sqrt {17}-16}{8}\)
The Correct Option is C
Solution and Explanation
![Let [t] denote the greatest integer less than or equal to t](https://images.collegedunia.com/public/qa/images/content/2024_01_18/Screenshot_1b4c1fc71705587192253.png)
\(∫_0^1 [ -8x^2 + 6x - 1] dx\)
\(= ∫_0^\frac 14 (-1)dx + ∫_{\frac 14}^{\frac 34} 0dx + ∫_{\frac 34}^{\frac 12} (-1)dx + ∫_{\frac {3+\sqrt {17}}{8}}^{\frac 34} (-1)dx + ∫_{\frac {3+\sqrt {17}}{8}}^{1}(-3)dx\)
\(= -\frac 14 -\frac 14 - 2 ( \frac {3+\sqrt {17}}{8} - \frac 34) - 3 (1 - \frac {3+\sqrt {17}}{8})\)
\(=\frac {\sqrt {17}-13}{8}\)
So, the correct option is (C): \(\frac {\sqrt {17}-13}{8}\)
Top Questions on Fundamental Theorem of Calculus
- If the slope of the tangent drawn at any point \((x, y)\) on the curve \(y = f(x)\) is \(y = 6x^2 + 10x - 9\) and \(f(2) = 0\), then \(f(-2) = \)?
- TS EAMCET - 2024
- Mathematics
- Fundamental Theorem of Calculus
$ \lim_{x \to -\frac{3}{2}} \frac{(4x^2 - 6x)(4x^2 + 6x + 9)}{\sqrt{2x - \sqrt{3}}} $
- TS EAMCET - 2024
- Mathematics
- Fundamental Theorem of Calculus
- The maximum interval in which the slopes of the tangents drawn to the curve \[ y = x^4 + 5x^3 + 9x^2 + 6x + 2 \] increase is:
- TS EAMCET - 2024
- Mathematics
- Fundamental Theorem of Calculus
If the function
$ f(x) = \begin{cases} \frac{\cos ax - \cos 9x}{x^2}, & \text{if } x \neq 0 \\ 16, & \text{if } x = 0 \end{cases} $
is continuous at $ x = 0 $, then $ a = ? $
- TS EAMCET - 2024
- Mathematics
- Fundamental Theorem of Calculus
- Let \( f : \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \rightarrow \mathbb{R} \) be a differentiable function such that \( f(0) = \frac{1}{2} \). If the \( \lim_{x \to 0} \frac{\int_{0}^{x} f(t) \, dt}{e^{x^2} - 1} = \alpha \), then \( 8\alpha^2 \) is equal to:
- JEE Main - 2024
- Mathematics
- Fundamental Theorem of Calculus
Questions Asked in JEE Main exam
- Which of the following quantity has the same dimensions as \( \sqrt{\frac{\mu_0}{\epsilon_0}} \)?
- JEE Main - 2025
- Units, Dimensions and Measurements
- Assume a living cell with 0.9%(\(w/w\)) of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data up to first decimal place only) The cell will:
- JEE Main - 2025
- osmosis
- Let $ \vec{a} = \hat{i} + 2 \hat{j} + \hat{k} $ and $ \vec{b} = 2 \hat{i} + \hat{j} - \hat{k} $. Let $ \hat{c} $ be a unit vector in the plane of the vectors $ \vec{a} $ and $ \vec{b} $ and perpendicular to $ \vec{a} $. Then such a vector $ \hat{c} $ is:
- JEE Main - 2025
- Vectors
- If $$ f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx, \quad f(0) = -6, \text{ then } f(1) \text{ is equal to:} $$
- JEE Main - 2025
- Continuity and differentiability
- 0.5 g of an organic compound gives 1.46 g CO\(_2\) and 0.9 g H\(_2\)O. Find the percentage composition of carbon.
- JEE Main - 2025
- Organic Chemistry
Concepts Used:
Fundamental Theorem of Calculus
Fundamental Theorem of Calculus is the theorem which states that differentiation and integration are opposite processes (or operations) of one another.
Calculus's fundamental theorem connects the notions of differentiating and integrating functions. The first portion of the theorem - the first fundamental theorem of calculus – asserts that by integrating f with a variable bound of integration, one of the antiderivatives (also known as an indefinite integral) of a function f, say F, can be derived. This implies the occurrence of antiderivatives for continuous functions.