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Solution:
We have,
Let
------ 1
So,,
------------ 2
Hence, by adding 1 and 2 ..
Solution:
We have,
Let, I=----- 1
So,------ 2
Adding 1 & 2 -------
Solution:
We have ,
Let,
So,
--------------- 2
Adding 1 and 2 --------
Solution:
Let,------ 1
---------- 2
Adding 1 & 2 -----
Solution:
Let,----------- (1)
So,
Solution:
---------- 1
----- 2
Adding 1 & 2 -------
Solution:
Let,
Let,
Now, x=0 ,, then
-------------------- 1
So,
------------- 2
Adding (1) and (2) ----------------
Solution:
Let,
then
------------ (1)
------------------- (2)
Adding (1) & (2) ------------
Solution:
Let,
Solution:
Let,
Equating coefficients, we get
So,
Solution:
------------- (1)
------------------ (2)
Adding (1) & (2) ----------------
Solution:
Let ,-------------- 1
So,
Let,
As, x=0, t=1 ; x=π , t=-1
Hence,
Solution:
Let,
Solution:
we have,
=
Since, f(x) = f(-x) , f(x) is an even function.
-------------- 1
---------------- 2
Adding 1 and 2 ----------------
Now, let
Putting 2x=t, we get
Solution:
Let,
Solution:
We have , I=\int\limits_0^{π} \frac{x}{1+cos\alpha sinx}dx ---------- 1
------- 2
Adding 1 and 2 ----
substituting s
when x=0 , t=0 ; x=π ,
Solution:
Let,
Solution:
Solution:
Solution:
Solution:
Now,
Let cosx=t
sinx dx=-dt
Solution:
------------------ 1
------------ 2
Adding 1 & 2 -------------
Let ,
Solution:
Let,
Here, f(x)=-f(x)
Hence, f(x) is odd function
Solution:
We have, is an even function.
Solution:
we have,
Since,
this is an odd function
Solution:
we have,
sin2x is even function
Hence,
Solution:
Solution:
we have ,
Let,
Then,
Solution:
Put cosx = t then -sinx dx = dt
Solution:
Let
is an odd function
Solution:
Solution:
Substitute π+x=u then dx=du
Solution:
Let,
Solution:
Applying the property ,
Thus,
-
Solution:
Let,
Solution:
let tanx = v
dv = sec2xdx
Solution:
Putthen
x=0 ⇒ t=0 and x=π ⇒
Solution:
we know,
Also here,
f(x) = f(2π -x)
So,
Solution:
then,
Solution:
We have ,
Then,
Let , 2a-t =x then dx=-dt
if t=a ⇒x=a
if t=2a ⇒ x=0
Hence Proved.
Solution:
We have,
Let 2a-t=x then dx=-dt
t=a , x=a ; t=2a , x=0
Solution:
we have ,
clearly f(x2) is an even function .
So,
Solution:
clearly , xf(x2) is odd function .
So,
Certainly. I'll provide a summary, 10 practice questions, and FAQs for Chapter 20: Definite Integrals, Exercise 20.4 Part B of RD Sharma's Class 12 mathematics textbook. Please note that this is based on general knowledge of the topic, as I don't have direct access to the specific textbook.
Summary:
Exercise 20.4 Part B of Chapter 20 likely focuses on more advanced techniques for evaluating definite integrals. This section probably covers:
1. Integration by parts for definite integrals
2. Definite integrals involving trigonometric functions
3. Evaluation of improper integrals
4. Application of substitution method in definite integrals
5. Definite integrals with logarithmic and exponential functions
10 Practice Questions:
1. Evaluate ∫[0 to π/2] x sin x dx using integration by parts.
2. Calculate ∫[0 to 1] x ln x dx.
3. Find the value of ∫[0 to ∞] e^(-x) dx.
4. Compute ∫[0 to π/4] sec^3 x dx.
5. Evaluate ∫[1 to e] (ln x)^2 dx.
6. Calculate ∫[0 to π/2] sin^3 x cos^2 x dx.
7. Find the value of ∫[0 to 1] x^n e^x dx, where n is a positive integer.
8. Evaluate ∫[0 to 1] √(1-x^2) dx.
9. Compute ∫[0 to π/2] sin x / (1 + cos x) dx.
10. Calculate ∫[1 to 2] (ln x) / x dx.
FAQs:
1. Q: What is integration by parts and when is it used?
A: Integration by parts is a technique used when integrating the product of two functions. It's particularly useful for integrals involving algebraic and transcendental functions.
2. Q: How do you evaluate improper integrals?
A: Improper integrals are evaluated by taking the limit of a definite integral as one or both of its limits approach infinity or a point where the integrand is undefined.
3. Q: What are some common trigonometric integrals to remember?
A: Some important ones include ∫ sin x dx = -cos x + C, ∫ cos x dx = sin x + C, and ∫ tan x dx = -ln|cos x| + C.
4. Q: How does the substitution method work for definite integrals?
A: When using substitution, you change both the integrand and the limits of integration. The new limits are found by applying the substitution to the original limits.
5. Q: What is the relationship between definite and indefinite integrals?
A: The Fundamental Theorem of Calculus connects them: if F(x) is an antiderivative of f(x), then ∫[a to b] f(x) dx = F(b) - F(a).
Would you like me to elaborate on any of these points or provide solutions to any of the practice questions?
Solution:
We have from LHS,
substituting
we get,
Solution:
------------------[ Given that f(a+b-x) = f(x) ]
Solution:
we have ,
Let, x=-t, then dx=-dt
x=-a ⇒ t=a
x=0 ⇒ t=0
Hence, Proved.
Solution:
Exercise 20.4 Part B of Chapter 20 likely focuses on more advanced techniques for evaluating definite integrals. This section probably covers: