The area under the curve $y = 2 x^{3} + 4 x^{2}$ between $x = 2, x = 4$ is

192.6

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198.6

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88.3

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172.3

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Solution

The correct option is A $192.6$
The area under the curve is given as $\int_{2}^{4} 2 x^{3} + 4 x^{2} d x \int_{2}^{4} 2 x^{3} d x + \int_{2}^{4} 4 x^{2} d x {2 \frac{x^{4}}{4} ∣ ∣ ∣}_{2}^{4} + {4 \frac{x^{3}}{3} ∣ ∣ ∣}_{2}^{4} 64 \times 2 - 8 + \frac{4^{4}}{3} - \frac{32}{3} 120 - \frac{224}{3} = 192.6$

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The area under the curve y=2x3+4x2 between x=2,x=4 is

The correct option is A 192.6The area under the curve is given as ∫422x3+4x2dx∫422x3dx+∫424x2dx2x44∣∣∣42+4x33∣∣∣4264×2−8+443−323120−2243=192.6

The area under the curve $y = 2 x^{3} + 4 x^{2}$ between $x = 2, x = 4$ is