Question

Find the area under the given curve over the indicated interval.\ y = x^2 + 3x + 2; [2, 6] The area is \frac{224}{3}. (Simplify your answer.)

          Find the area under the given curve over the indicated interval.\
y = x^2 + 3x + 2; [2, 6]
The area is \frac{224}{3}.
(Simplify your answer.)

Find the area under the given curve over the indicated interval.y = x^2 + 3x + 2; [2, 6]
The area is (224)/(3).
(Simplify your answer.)

Added by Josefina G.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Steven Clarke

04/15/2022

Step 1

The antiderivative of x^2 is (1/3)x^3, the antiderivative of 3x is (3/2)x^2, and the antiderivative of 2 is 2x. Therefore, the antiderivative of y=x^2+3x+2 is (1/3)x^3 + (3/2)x^2 + 2x. Show more…

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Find the area under the given curve over the indicated interval. y=x^(2)+3x+2;[2,6] The area is (224)/(3). (Simplify your answer.) Find the area under the given curve over the indicated interval. KK y=x2+3x+2; [2, 6] 224 The area is 3 (Simplify your answer.)