Sol. Since f(t) = is continuous, therefore g'(x) =
Ex.2 If F(t) = dx, find F'(1), F'(2), and F'(x).
Sol. The integrand in this example is the continuous function f defined by f(x) =
Ex.3 Find
Sol. Let u = x^{4}. Then
Ex.4 FInd the derivative of F(x) =
Sol.
= (cos u) (3x^{2}) = (cos x^{3}) (3x^{2})
F'(x) = (cos x^{3}) (3x^{2}).
Ex.5 Let f(x) = . 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^{2 }+ x + 1)^{2} – (a + 1) (x^{2} + x + 1) (x^{2} – x + 1).
Now, f'(x) = 0 ⇒ (a – 1) (x^{2} + x + 1) – (a + 1) (x^{2} – x + 1) = 0 ⇒ x^{2} – ax + 1 = 0.
For distinct real roots D > 0 i.e. a^{2} – 4 > 0 ⇒ a^{2} > 4 ⇒ a ∈ (∝, 2) U (2, ∝)
Ex.6 Show that for a differentiable function f(x),
(where [ * ] denotes the greaetest integer function and n ε N)
Sol.
= – f(1) – f(2) – ........ – f(n – 1) – f(n)
Ex.7 Evaluate
Sol.
Ex.8 Evaluate
Sol.
We must now evaluate the integrals on the right side separately :
Since both of these integrals are convergent, the given integral is convergent and Since 1/(1 + x^{2}) > 0, the given improper integral can be interpreated as the area of the infinite region that lies under the curve y = 1/(1 + x^{2}) and above the xaxis (see Figure).
Ex.9 Find
Sol.
We note first that the given integral is improper because f(x) = 1/√(x2) has the vertical asymptote x = 2. Since the infinite discontinuity occurs at the left end point of [2, 5]
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.10 Evaluate
Sol. We know that the function f(x) = ln x has a vertical asymptote at 0 since Thus, the given integral is improper and we have
Now we integrate by parts with u = ln x, dv = dx, du = dx/x, and v = x
=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 :
Therefore = –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 xaxis is 1.
Ex.11 Evaluate (where [ * ] denotes the greatest integer function)
Sol.
for x > ln 2 ⇒ e^{x} > 2 ⇒ e^{x} < 1/2 ⇒ 2e^{–x }< 1 ∴ 0 ≤ 2e^{x} < 1 [2e^{x}] = 0
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1. What is the Leibniz Rule for the derivative of an antiderivative? 
2. How is the Leibniz Rule derived? 
3. Can the Leibniz Rule be applied to all functions? 
4. How is the Leibniz Rule used in practice? 
5. Are there any limitations to the Leibniz Rule? 

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