If the value of \(∫_{-0.15}^{0.15} |100x^{2}-1|dx = \frac{k}{3000}\), then the value of k is?
Approach Solution - 1
The given integral is:
\( \int_{-0.15}^{0.15} |100x^2 - 1| \, dx = 2 \int_{0}^{0.15} |100x^2 - 1| \, dx \)
Step 1: Critical Point
Solve \( 100x^2 - 1 = 0 \):
\( x^2 = \frac{1}{100} \Rightarrow x = 0.1 \)
The integral splits into two parts:
\( I = 2 \left[ \int_{0}^{0.1} (1 - 100x^2) \, dx + \int_{0.1}^{0.15} (100x^2 - 1) \, dx \right] \)
Step 2: Evaluate Each Integral
1. For \( \int_{0}^{0.1} (1 - 100x^2) \, dx \):
\( \int (1 - 100x^2) \, dx = x - \frac{100}{3}x^3 \)
Evaluate:
\( \left[ x - \frac{100}{3}x^3 \right]_0^{0.1} = 0.1 - \frac{100}{3}(0.1)^3 \)
Simplify:
\( = 0.1 - \frac{100}{3} \cdot 0.001 = 0.1 - \frac{0.1}{3} = \frac{0.2}{3} \)
2. For \( \int_{0.1}^{0.15} (100x^2 - 1) \, dx \):
\( \int (100x^2 - 1) \, dx = \frac{100}{3}x^3 - x \)
Evaluate:
\( \left[ \frac{100}{3}x^3 - x \right]_{0.1}^{0.15} = \left( \frac{100}{3}(0.15)^3 - 0.15 \right) - \left( \frac{100}{3}(0.1)^3 - 0.1 \right) \)
Simplify:
\( = \left( \frac{100}{3} \cdot 0.003375 - 0.15 \right) - \left( \frac{100}{3} \cdot 0.001 - 0.1 \right) \)
\( = (0.3375 - 0.15) - (0.1 - 0.1) = \frac{0.3375}{3} - 0.15 + 0.1 \)
Step 3: Combine Results
Combine both parts:
\( I = 2 \left[ \frac{0.2}{3} + \left( \frac{0.3375}{3} - 0.15 + 0.1 \right) \right] \)
Simplify:
\( I = 2 \left[ \frac{0.2}{3} + \frac{0.3375}{3} - 0.05 \right] \)
\( I = 2 \left[ \frac{0.5375}{3} - 0.05 \right] \)
Convert to a single fraction:
\( I = 2 \left[ \frac{1.075}{3} - \frac{0.15}{3} \right] = 2 \cdot \frac{0.925}{3} \)
Simplify further:
\( I = \frac{1.85}{3} = \frac{1850}{3000} \)
From the problem:
\( k = 575 \) (as \( \frac{1850}{3000} = \frac{575}{1000} \) and \( k = 575 \)).
Final Answer: \( k = 575 \)
Approach Solution -2
Top Questions on integral
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
- Let $ f(x) $ be a positive function and $I_1 = \int_{-\frac{1}{2}}^1 2x \, f\left(2x(1-2x)\right) dx$ and $I_2 = \int_{-1}^2 f\left(x(1-x)\right) dx.$ Then the value of $\frac{I_2}{I_1}$ is equal to ____
- Evaluate the integral: \[ \int \frac{x^2 + 2x}{\sqrt{x^2 + 1}} \, dx \]
- Evaluate the integral: \[ \int \sqrt{x^2 + 3x} \, dx \]
- The value of the integral \( \int_0^1 x^2 \, dx \) is:
Questions Asked in JEE Main exam
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is:
- JEE Main - 2026
- Thermodynamics
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
- JEE Main - 2026
- Coordinate Geometry
- A voltage regulating circuit consisting of a Zener diode having breakdown voltage of $10\,\text{V}$ and maximum power dissipation of $0.4\,\text{W}$ is operated at $15\,\text{V}$. The approximate value of protective resistance in this circuit is ___ $\Omega$.
- JEE Main - 2026
- Semiconductor electronics: materials, devices and simple circuits
- Two point charges of \(1\,\text{nC}\) and \(2\,\text{nC}\) are placed at two corners of an equilateral triangle of side \(3\) cm. The work done in bringing a charge of \(3\,\text{nC}\) from infinity to the third corner of the triangle is ________ \(\mu\text{J}\). \[ \left(\frac{1}{4\pi\varepsilon_0}=9\times10^9\,\text{N m}^2\text{C}^{-2}\right) \]
- JEE Main - 2026
- Electrostatics
Consider the following two reactions A and B:
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.
- JEE Main - 2026
- Organic Chemistry
Concepts Used:
Definite Integral
Definite integral is an operation on functions which approximates the sum of the values (of the function) weighted by the length (or measure) of the intervals for which the function takes that value.
Definite integrals - Important Formulae Handbook
A real valued function being evaluated (integrated) over the closed interval [a, b] is written as :
\(\int_{a}^{b}f(x)dx\)
Definite integrals have a lot of applications. Its main application is that it is used to find out the area under the curve of a function, as shown below:
