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The area bounded by the curve y = 2ˣ, x – axis, ordinates x = 0 and x = 4 is
To determine the area bounded by the curve y = 2ˣ, the x-axis, and the ordinates x = 0 and x = 4, we are required to compute the definite integral of 2ˣ from x = 0 to x = 4. Step 1: Write down the integral The area is given by: A = ∫₀⁴ 2ˣ dx Step 2: Evaluate the integral The integral of 2ˣ is: ∫ 2ˣRead more
To determine the area bounded by the curve y = 2ˣ, the x-axis, and the ordinates x = 0 and x = 4, we are required to compute the definite integral of 2ˣ from x = 0 to x = 4.
Step 1: Write down the integral
The area is given by:
A = ∫₀⁴ 2ˣ dx
Step 2: Evaluate the integral
The integral of 2ˣ is:
∫ 2ˣ dx = (2ˣ) / ln 2
Now, calculate the area under the curve from x = 0 to x = 4:
A = [(2ˣ) / ln 2]₀⁴
At x = 4:
(2⁴) / ln 2 = 16 / ln 2
At x = 0:
(2⁰) / ln 2 = 1 / ln 2
Therefore, the area:
A = (16 / ln 2) – (1 / ln 2) = 15 / ln 2
Step 3: Final result
So the region is:
A = 15 / ln 2 square units
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Which of the following is the correct order of inter particle spaces present in the matter? (a) Solid and Liquid and Gas (b) Solid and Liquid Liquid and Gas (d) Solid Gas
(b) Interparticle space (space between two neighbouring particles) is maximum in a gas because there is least force of attraction and there is maximum freedom. https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-1/
(b) Interparticle space (space between two neighbouring particles) is maximum in a gas because there is least force of attraction and there is maximum freedom.
https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-1/
See lessTwo beakers A and B contain normal water and hot water respectively. Mohan adds equal amount of blue copper sulphate crystals to them. Which of the following observations is correct? (a) Blue colour will spread more quickly in A. (b) Blue colour will spread more quickly in B. (c) Blue colour will take the same time in both A and B to spread. (d) Blue colour will not spread in either of the beakers.
(b) With increase in temperature, the movement of particles increases. https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-1/
(b) With increase in temperature, the movement of particles increases.
https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-1/
See lessAnti-derivative of (tan x – 1)/(tan x + 1) with respect to x is
To determine the antiderivative (indefinite integral) of I = ∫ (tan x - 1) / (tan x + 1) dx Step 1: Substituting x in terms of a trigonometric identity We apply the identity : tan(A - B) = (tan A - tan B) / (1 + tan A tan B) Here, we choose A = π/4 and B = x, so tan(π/4 - x) = (tan(π/4) - tan x) / (Read more
To determine the antiderivative (indefinite integral) of
I = ∫ (tan x – 1) / (tan x + 1) dx
Step 1: Substituting x in terms of a trigonometric identity
We apply the identity :
tan(A – B) = (tan A – tan B) / (1 + tan A tan B)
Here, we choose A = π/4 and B = x, so
tan(π/4 – x) = (tan(π/4) – tan x) / (1 + tan(π/4) tan x)
Since tan(π/4) = 1, this reduces to:
tan(π/4 – x) = (1 – tan x) / (1 + tan x)
Taking the negative,
– tan(π/4 – x) = (tan x – 1) / (tan x + 1)
So, our integral is:
I = ∫ – tan(π/4 – x) dx
Step 2: Finding the Integral
We know:
∫ tan u du = log | sec u | + C
Substituting u = π/4 – x, we get:
I = – ∫ tan(π/4 – x) dx
= – log | sec(π/4 – x) | + C
Conclusion
Therefore, the right answer is: – log | sec(π/4 – x) | + C
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In a sphere the rate of change of volume is
The correct answer is: surface area times the rate of change of radius. This follows because the volume V of a sphere is connected to its radius r by the formula: V = (4/3) π r³ The rate of change of volume with respect to time is: dV/dt = 4 π r² (dr/dt) Here, 4 π r² is the surface area of the spherRead more
The correct answer is: surface area times the rate of change of radius.
This follows because the volume V of a sphere is connected to its radius r by the formula:
V = (4/3) π r³
The rate of change of volume with respect to time is:
dV/dt = 4 π r² (dr/dt)
Here, 4 π r² is the surface area of the sphere, and (dr/dt) is the rate of change of the radius. Therefore, the rate of change of volume is equal to the surface area times the rate of change of radius.
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