Find the area of the region bounded by the x -axis and the graph of the function f(x)=(x)/((1+x^

Find the area of the region bounded by the x -axis and the...

Key Concepts

Properties of Even and Odd Functions

Recognizing whether a function is even or odd can simplify calculations, particularly when the interval of interest is symmetric about the origin. For odd functions, integration over symmetric limits results in cancellation of positive and negative areas, so it is important to use absolute values to correctly compute the total area between the curve and the x-axis.

Substitution Method in Integration

This technique, also known as u-substitution, is used to simplify integrals by transforming the integrand into a more manageable form. It is particularly useful when dealing with composite functions or rational expressions, such as those that arise in the integrals of functions defined by complicated algebraic forms.

Definite Integration

This concept involves computing the area under a curve by setting up and evaluating a definite integral over a specified interval. It is fundamental for finding the area bounded by a graph and the x-axis, especially when the region spans segments where the function may change sign.

Transcript

00:01 In the question given graph of function fx that is equal to x divided by 1 plus x square whole square.

00:13 Now in the question it is asking to find the area which is bounded by x -axis and this given graph of function fx.

00:27 So by using these two data i have to find area of reason.

00:39 So for finding this first i will write area in term of integration by using these two data.

00:46 So i can say that area will be equal to integral as you see the limits will be at minus infinity to plus infinity because these limits are x limits and it is saying that it is bounded on x -axis by bounded by x -axis so i can say that minus infinity plus infinity to plus infinity integral f x d x is x divided by one plus x square mole square d x now next step is i am using a property which is that if minus infinity plus infinity fx dx is given then it will be written as like minus infinity to c integral fx dx plus c to minus infinity plus infinity integral fx dx so using this property i will write area in term of that area is equal to let assume c is equal to 0 so use this property so i can say that area will be equal to minus infinity to c is equal to 0 so it will be 0 fx means x divided by 1 plus x square whole square d x plus 0 to plus infinity f x d x so fx d x so fx will be written as that x divided by 1 plus x square whole square d x now next step is again it will return in other terms by using the property of function that i can say that area will be equal to limit a tends to sorry a tends to minus infinity integral a to zero x times of x divided by x is 1 plus x square whole square d x plus limit b tends to infinite 0 to b x divided by 1 plus x square all square d x now next step is to find the value of integration of x divided by 1 plus x square all squared d x for finding for finding this i will use the substitution method so before applying the substitution method let assume that you is equal to 1 plus x is square so du will be that will be equal to after taking differentiation on both side so i can say that i can say that do is equal to 2x d x now i will try to find the value of integral x divided by 1 plus x square all square d x because during for 5 because when i am writing to find the area there is you see i have to find the value of integration of x divided by 1 plus x square all square d x so finding this i will try to separate put get the value of this integral by applying the substitution substitution method so i can say that 1 plus x square so as you see in the numerator there is one plus so denominator there is one plus x squared whole square so it will be written as that 1 plus x square will be u u square and x dx will be that will be du divided by 2 integration so i have to put this value of integration of x divided by 1 plus x square all squared d x in the line of area so putting there the limits also changed as you see the limits because u is equal to 1 plus x square so limits i will try to first find the limits of this first term and this second term so limits are that as you see in first term it is saying that the lower limit of axis that is equal to a then the upper then the value of lower limit of u that will be equal to as you see here that where a will be tends to minus infinity it means lower limit is that x tends to minus infinity so you will be so lower so lower so lower limit of you will be that is equal to minus infinity and in the case in the first term it is you see that upper limit of a axis equal to 0 so upper limit of u will be that it will be by using as assuming that u is equal to 1 plus x square so put the value of x that is 0 then i will get the value of upper limit that is 0 plus 1 means 1 so for first term the lower limit is minus infinity and the upper limit is 1 now for second term i will find the limits.

07:22 So as you see, the lower limit of x is that is equal to 0, then i can say that lower limit of u will be 1.

07:31 And again, in the second term, the upper limit is given that is x tends to b, means tends to positive you...

Please give Ace some feedback