Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Mathematics (Maths) Class 12

JEE : Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

The document Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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Derivative of Antiderivative (Leibnitz Rule) 
If h(x) & g(x) are differentiable function of x then, Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev


Ex.15 Find the derivative of the function g(x) = Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Sol. Since f(t) = Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev   is continuous, therefore g'(x) = Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Ex.16 If F(t) = Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRevdx, find F'(1), F'(2), and F'(x).

Sol. The integrand in this example is the continuous function f defined by f(x) = Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev


Ex.17 Find Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Sol. Let u = x4. Then

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Ex.18 FInd the derivative of F(x) = Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Sol.

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev = (cos u) (3x2) = (cos x3) (3x2)

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

F'(x) = (cos x3) (3x2).

Ex.19 Let f(x) = Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev. Find the value of 'a' for which f'(x) = 0 has two distinct real roots.

Sol. Differentiating the given equation, we get f'(x) = (a – 1) (x+ x + 1)2 – (a + 1) (x2 + x + 1) (x2 – x + 1).
Now, f'(x) = 0 ⇒ (a – 1) (x2 + x + 1) – (a + 1) (x2 – x + 1) = 0 ⇒ x2 – ax + 1 = 0.
For distinct real roots D > 0 i.e. a2 – 4 > 0 ⇒  a2 > 4  ⇒  a ∈ (-∝, -2) U (2, ∝)


Ex.20 Show that for a differentiable function f(x), 

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev
(where [ * ] denotes the greaetest integer function and n ε N)

Sol.

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

= – f(1) – f(2) – ........ – f(n – 1) – f(n) Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Ex.21 Evaluate Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev


Sol.

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Ex.22 Evaluate Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Sol.

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev
We must now evaluate the integrals on the right side separately :

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev 

Since both of these integrals are convergent, the given integral is convergent and  Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev Since 1/(1 + x2) > 0, the given improper integral can be interpreated as the area of the infinite region that lies under the curve y = 1/(1 + x2) and above the x-axis (see Figure). 

Ex.23 Find Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Sol.

We note first that the given integral is improper because f(x) = 1/√(x-2) has the vertical asymptote x = 2. Since the infinite discontinuity occurs at the left end point of [2, 5] 

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Thus, the given improper integrat is convergent and, since the integrand is positive, we can interpret the value of the integral as the area of the shaded region in Figure.

Ex.24 Evaluate  Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Sol. We know that the function f(x) = ln x has a vertical asymptote at 0 since Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev Thus, the given integral is improper and we have  Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Now we integrate by parts with u = ln x, dv = dx, du = dx/x, and v = x

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev =1 ln – t ln t – (1 – t)  = – t ln t – 1 + t

To find the limit of the first term we use I'Hopital's Rule :

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Therefore   Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev  = –0 – 1 + 0 = –1

Figure shows the geometric interpretation of this result. The area of the shaded region above y = ln x and below the x-axis is 1.


Ex.25 Evaluate  Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev   (where [ * ] denotes the greatest integer function)

Sol.

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

for x > ln 2  ⇒  ex > 2 ⇒ e-x < 1/2 ⇒ 2e–x < 1  ∴ 0 ≤ 2e-x < 1 [2e-x] = 0

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev

 

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