∫ab (x−a)(b−x)dx(b>a) is equal to

Let t=x−a+x−b2=x−12(a+b)and α=b−a2 , sot that∫ab (x−a)(b−x)dx=∫-αα (t+α)(α−t)dt=2∫0α α2−t2dt=tα2−t2+α2sin−1⁡tα0α=πα22=π(b−a)28.

∫ab (x−a)(b−x)dx(b>a) is equal to

Let t=x−a+x−b2=x−12(a+b)and α=b−a2 , sot that∫ab (x−a)(b−x)dx=∫-αα (t+α)(α−t)dt=2∫0α α2−t2dt=tα2−t2+α2sin−1⁡tα0α=πα22=π(b−a)28.

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\int_{a}^{b} \sqrt{(x - a) (b - x)} d x (b > a)$ is equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$π (b - a)^{2} / 8$

$π (b + a)^{2} / 8$

$π (b - a)^{2}$

$π (b + a)^{2}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let $t = \frac{x - a + x - b}{2} = x - \frac{1}{2} (a + b)$

and $α = \frac{b - a}{2}$ , sot that

$\begin{array}{l} \int_{a}^{b} \sqrt{(x - a) (b - x)} d x = \int_{- α}^{α} \sqrt{(t + α) (α - t)} d t \\ = 2 \int_{0}^{α} \sqrt{α^{2} - t^{2}} d t \\ {= t \sqrt{α^{2} - t^{2}} + α^{2} \sin^{- 1} (\frac{t}{α})]}_{0}^{α} \\ = \frac{π α^{2}}{2} = \frac{π (b - a)^{2}}{8} . \end{array}$