Given;The curve x2 = 4y and the straight line x = 4y – 2By substituting for 4y;⇒ x = x2 − 2⇒ x2 − x − 2 = 0⇒ (x + 1) (x − 2) = 0∴ x = −1, 2.Required area

Q. 254.2( 6 Votes )

The area of the r A. sq units

B. sq units

C. sq units

D. sq units

Class 12th Mathematics - Exemplar 8. Application of Integrals

Answer :

Given;

The curve x² = 4y and the straight line x = 4y – 2

By substituting for 4y;

⇒ x = x² − 2

⇒ x² − x − 2 = 0

⇒ (x + 1) (x − 2) = 0

∴ x = −1, 2.

Required area

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The area of the r A. sq unitsB. sq unitsC. sq unitsD. sq units

The area of the r A. sq units

B. sq units

C. sq units

D. sq units