(x-a)(x-b) can be written as x2 - (a+b)x + ab.
Therefore,
x2- (a+b)x + ab
= x2- (a+b)x + \(\frac {(a+b)^2}{4}\) - \(\frac {(a+b)^2}{4}\) + ab
= [x-(\(\frac {a+b}{2}\))]2 - \(\frac {(a-b)^2}{4}\)
⇒ \(∫\)\(\frac {1}{\sqrt {(x-a)(x-b)}}\ dx\) = \(∫\frac {1}{\sqrt {{x-(\frac {a+b}{2})}^2-(\frac {a-b}{2})^2}} dx\)
Let x - (\(\frac {a+b}{2}\)) = t
∴ dx = dt
⇒ \(∫\frac {1}{\sqrt {{x-(\frac {a+b}{2})}^2-(\frac {a-b}{2})^2}} dx\) = \(∫\frac {1}{\sqrt {t^2-(\frac {a-b}{2})^2}}dt\)
= \(log \ |t+\sqrt {t^2-(\frac {a-b}{2})^2|}+C\)
= \(log \ |{x-(\frac {a+b}{2})}+\sqrt {(x-a)(x-b)}|+C\)
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Find the area of the region bounded by the curve y2=x and the lines x=1,x=4 and the x-axis
Find the area of the region bounded by y2=9x, x=2, x=4 and the x-axis in the first quadrant.
Find the area of the region bounded by x2=4y,y=2,y=4 and the x-axis in the first quadrant.
Find the area of the region bounded by the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\)
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
These are tabulated below along with the meaning of each part.
