Question

Integrate with respect to x :
∫10(2x2−4x−6)

a 10/4

b 2/3

c 14/3

d 5/6

Answer 1

Answer on Question #50920, Math, Integral Calculus

Integrate with respect to $x$:

$$\int_{0}^{1} (2x^2 - 4x - 6) \, dx$$

a 10/4

b 2/3

c 14/3

d 5/6

Solution

$$\int_{0}^{1} (2x^2 - 4x - 6) \, dx = \left(\frac{2x^3}{3} - 2x^2 - 6x\right)\bigg|_{0}^{1} = \left(\frac{2 \cdot 1^3}{3} - 2 \cdot 1^2 - 6 \cdot 1\right) - \left(\frac{2 \cdot 0^3}{3} - 2 \cdot 0^2 - 6 \cdot 0\right) = -\frac{22}{3}$$

Maybe, condition has mistake

Suppose, $\int_{0}^{1} (2x^2 - 4x + 6) \, dx$, then

$$\int_{0}^{1} (2x^2 - 4x + 6) \, dx = \left(\frac{2x^3}{3} - 2x^2 + 6x\right)\bigg|_{0}^{1} = \left(\frac{2 \cdot 1^3}{3} - 2 \cdot 1^2 + 6 \cdot 1\right) - \left(\frac{2 \cdot 0^3}{3} - 2 \cdot 0^2 + 6 \cdot 0\right) = \frac{14}{3}$$

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