Applications Of Integrals

y=(x²-1)² Minimum occurs when x²=1 x=±1. Only x=1 lies in interval. yᵢₙ=(1-1)²=0

f'(x)= x+ x-a For decreasing function: f'(x)\le0 x+ x a Maximum value of x+ x=√(2). Thus: a√(2)

Step 1: Function is odd and symmetric.
Step 2: Area=2₀¹ x² dx=(2)/(3)

Step 1: Area integral. ₀π/⁴tan xdx
Step 2: Evaluation. =- x|₀π/⁴=(\ln2)/(2)

Step 1: Integrate to find the area.
The area under the curve is given by: Thus, the correct answer is (3).

Step 1: Set up the area integral.
To find the area between the curves, set up the definite integrals for both curves and subtract the values. Use the limits from to .
Step 2: Calculate the area.
After solving the integrals, the area is found to be . Final Answer:

Step 1:
Step 2: Differentiate w.r.t. :

Step 1: Condition implies region between lines .
Step 2: In polar coordinates, this corresponds to angles:
Step 3: Area inside unit circle:

Vertex of the parabola is a point which lies on the axis of the parabola, which is a line to the directrix through the focus, i.e., and equidistant from the focus and directrix , so that the vertex is .

Split at :

The condition corresponds to angles between
and similarly in all four quadrants.

Total angular region:


Area inside the unit circle:

Step 1: Understanding the Concept
The area between two curves is found by integrating the difference between the "right" curve and the "left" curve (if integrating with respect to ) over the interval defined by their points of intersection.

Step 2: Finding Points of Intersection
Set the two equations equal to each other: The curves intersect at and .

Step 3: Calculating the Area
In the interval , the line is to the right of the parabola .

Step 4: Final Answer
The area of the region is .