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The value of the definite integra equal to a.
Let a, b, c be nonzero real numbers such that
Then the quadratic equation ax^{2} + bx +c= 0 has
Now we know that if then it means that
f (x) is + ve on some part of (α, β) and – ve on other part of (α, β).
But here 1 + cos^{8} x is always + ve,
∴ ax^{2} + bx + c is + ve on some part of [1 , 2] and – ve on other part [1, 2]
∴ ax^{2} + bx + c= 0 has at least one root in (1, 2).
⇒ ax^{2} + bx + c = 0 has at least one root in (0,2).
The area bounded by the curves y = f(x), the xaxis and the ordinates x = 1 and x = b is (b – 1) sin (3b + 4). Then f(x) is
Differentiating both sides w.r.t b, we get
f(b) = 3(b – 1) cos(3b + 4) + sin(3b + 4)
⇒ f(x) = 3(x – 1) cos(3x + 4) + sin(3x + 4)
The value of the integral
For any integer n the integral ––
has the value
Let f : R → R and g : R → R be continous functions. Then the value of the integral
We have,
Let F (x) = (f (x) + f ( x)) (g (x)  g (x))
then F (x) = (f ( x) + f (x))(g ( x) g (x)) =  [f (x) + f ( x)][ g(x)  g (x)]
=F (x)
∴ F(x) is an odd function, ∴ we get I = 0
The value of
If f (x) and then constants A and B are
The value of represents the greatest integer function is
equals
We have
Adding (1) and (2), we get
If for a real number y, [y] is the greatest integer less than or equal to y, then the value of the integral
In the range we have to find the value of
where f is such that
Then g(2) satisfies the inequality
(applying line integral on inequality)
… (1)
(applying line integral on inequality)
… (2)
From (1) and (2), we get
[∵ e^{cos x} sin x is an odd function.]
The value of the integral
We know that for and hence
and for
The value of
.....(1)
Adding (1) and (2),
(even function)
...(3)
....(4)
Adding (3) and (4)
∴ I = π /2
The area bounded by the curves y = x –1 and y = –x + 1 is
The given lines are
y = x – 1; y = – x – 1;
y = x + 1 and y = – x + 1
which are two pairs of parallel lines and distance between the lines of each pair is √2 Also non parallel lines are perpendicular. Thus lines represents a square of side √2 Hence, area = (√2)^{2} = 2 sq. units.
Then the real roots of the equation x^{2} – f '(x) = 0 are
Now the given equation x^{2}  f ' (x)=0 becomes
Let T > 0 be a fixed real number. Suppose f is a continuous function such that for all
Given that T > 0 is a fixed real number. f is continuous
⇒ f is a periodic function of period T
The integral
then the expression for l(m, n) in terms of l(m + 1, n – 1) is
Intergrating by parts considering (1 + t)^{n} as first function, we get
increases in
∴ f(x) increases when x < 0
The area bounded by the curves and xaxis in the 1^{st} quadrant is
The curves given are
and xaxis y = 0 ...(3)
Eqn. (1), [y^{2} = x] represents right handed parabola but with +ve values of y i.e., part of curve lying above xaxis.
Solving (1) and (2) we get,
Also (2) meets xaxis at (3, 0)
Shaded area is the required area given by
If f (x) is differentiable and equals
Differentiating both sides w.r.t. t
[Using Leibnitz theorem]
The value of the integral
The area enclosed between the curves y = ax^{2} and x = ay^{2} (a > 0) is 1 sq. unit, then the value of a is
y = ax^{2} and x = ay^{2}
Points of intersection are O (0, 0) and
Th e area bounded by the par abolas y = (x + 1)^{2} and y = (x – 1)^{2} and the line y = 1/4 is
The given curves are
y = (x+ 1)^{2} ...(1)
upward parabola with vertex at (–1 ,0) meeting y –axis at (0, 1)
y = (x 1)^{2} ...(2)
a line parallel to x–axis meeting (1) at
and meeting (2) at
The graph is as shown
The required area is the shaded portion given by ar (BPCQB) = 2 Ar(PQCP) (by symmetry)
The area of the region between the curves and bounded by the lines x = 0 and is
The given curves are
∴ The area bounded by the above curves, by the lines
Also when
Let f be a nonnegative function defined on the interval
and f (0) = 0, then
Given that f is a non negative function defined on
Differentiating both sides with respect to x, we get
Integrating both sides with respect to x, we get
∵ Given that f (0) = 0 ⇒ C = 0
Hence f (x) = ± sin x
But as f (x) is a non negative function on [0, 1]
∴ f (x) = sinx.
The value of
Applying L’ Hospital’s rule, we get
Let f be a realvalued function defined on the interval (–1, 1) such that and let f ^{–1} be the inverse function of f. Then (f –1)' (2) is equal to
Now on differentiating, we get
The value of
Adding values of I in equation (1) and (2)
Let the straight lin e x = b divide the area enclosed by y = (1 – x)^{2} , y = 0, and x = 0 into two parts R_{1} (0 < x < b) and R_{2} (b < x < 1) such that R_{1}  R_{2} = 1/4. Then b equals
Let f : [– 1, 2] → [0, ∞) be a continuous function such that f (x) = f (1 – x) for all x ∈ [–1, 2]
and R_{2} be the area of the region bounded by y = f (x), x = –1, x = 2, and the xaxis.
Then
We have
The value of the integral
[as x^{2} cos x is an even cos x is an odd function]
The area enclosed by the curves y = sin x + cos x and y = cosx  sinx over the interval
The rough graph of y = sin x + cos x and y = cos x – sin x suggest the required area is
(the set of all real number) be a positive, nonconstant and differentiable function such that Then the value of in the interval
⇒ e^{2 x} f (x) is strictly decreasing function on
Also given that f(x) is positive function so f(x) > 0
The following integral
The value of is equal to
Area of the region is equal to
Solving (i) and (ii), we get intersection points as (1, 2), (6, 3), (– 4, 1), (–39, –6)
The graph of given region is as follows
Required area = Area (trap PQRS) – Area (PST + TQR)
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