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Evaluate [tex]\int_{-2}^{3} 7x^6 \, dx[/tex] using the fundamental theorem of calculus.

A. [tex]\int_{-2}^{3} 7x^6 \, dx = 4655[/tex]

B. [tex]\int_{-2}^{3} 7x^6 \, dx = -2315[/tex]

C. [tex]\int_{-2}^{3} 7x^6 \, dx = 2315[/tex]

D. [tex]\int_{-2}^{3} 7x^6 \, dx = -4655[/tex]

Answer :

The correct evaluation of ∫−237x⁶dx using the fundamental theorem of calculus is (-237/7)x⁷, which is none of the options provided.

To evaluate the integral ∫−237x⁶dx using the fundamental theorem of calculus, we need to follow these steps:

Identify the antiderivative of the integrand. In this case, we have -237x⁶. To find its antiderivative, add 1 to the exponent and divide by the new exponent.

The antiderivative of xⁿ is (1/(n+1))xⁿ⁺¹. So, the antiderivative of -237x⁶ is (-237/(6+1))x⁶⁺¹= -237/7x⁷.

Apply the fundamental theorem of calculus. The theorem states that if F(x) is the antiderivative of f(x), then ∫[a to b]f(x)dx = F(b) - F(a). In our case, a and b are not provided, so we can omit this step.

Evaluate the antiderivative at the upper and lower limits of integration. Since no limits are given, we can skip this step as well.

Simplify the antiderivative. In our case, the antiderivative is (-237/7)x⁷.

Therefore, the correct evaluation of ∫−237x⁶dx using the fundamental theorem of calculus is (-237/7)x⁷, which is none of the options provided.

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