Evaluate int_(0)^(6)(7e^(0.5 x)-5)dx and express the answer in simplest form. Answer:

Identify integral: Identify the integral to be solved. We need to evaluate the integral of the function 7e^(0.5x) - 5 from 0 to 6. This can be written as: ∫(7e^(0.5x) - 5)dx from 0 to 6.Break into two: Break the integral into two separate integrals. The integral of a sum is the sum of the integrals, so we can write: ∫(7e^(0.5x) - 5)dx = ∫7e^(0.5x)dx - ∫5dx from 0 to 6.Evaluate first integral: Evaluate the first integral ∫7e^(0.5x)dx from 0 to 6. The antiderivative of e^(0.5x) is (2e^(0.5x)), because when we differentiate (2e^(0.5x)), we get e^(0.5x) times the derivative of (0.5x) which is 0.5, and 2 * 0.5 = 1, giving us back e^(0.5x). Therefore, the antiderivative of 7e^(0.5x) is 7 * 2e^(0.5x) = 14e^(0.5x).Evaluate second integral: Evaluate the second integral ∫5dx from 0 to 6. The antiderivative of a constant 5 is 5x. So the integral of 5 from 0 to 6 is 5x evaluated from 0 to 6.Combine and evaluate: Combine the antiderivatives and evaluate from 0 to 6. We have the antiderivatives 14e^(0.5x) and 5x. Now we need to evaluate these from 0 to 6: (14e^(0.5x) - 5x) | from 0 to 6 = (14e^(0.5*6) - 5*6) - (14e^(0.5*0) - 5*0).Perform final evaluation: Perform the evaluation using the bounds 0 and 6. First, we evaluate at the upper bound x = 6: 14e^(0.5*6) - 5*6 = 14e^3 - 30. Next, we evaluate at the lower bound x = 0: 14e^(0.5*0) - 5*0 = 14e^0 - 0 = 14*1 - 0 = 14. Now, subtract the lower bound evaluation from the upper bound evaluation: (14e^3 - 30) - (14) = 14e^3 - 30 - 14 = 14e^3 - 44.

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Evaluate
int_(0)^(6)(7e^(0.5 x)-5)dx and express the answer in simplest form.
Answer:

Evaluate $\int_{0}^{6}\left(7 e^{0.5 x}-5\right) d x$ and express the answer in simplest form. $\newline$ Answer:

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Calculus

Evaluate definite integrals using the power rule

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Q. Evaluate

\int_{0}^{6}\left(7 e^{0.5 x}-5\right) d x

and express the answer in simplest form.

\newline

Answer:

Identify integral: Identify the integral to be solved. $\newline$ We need to evaluate the integral of the function $7e^{0.5x} - 5$ from $0$ to $6$ . This can be written as: $\newline$ $\int_{0}^{6}(7e^{0.5x} - 5)\,dx.$
Break into two: Break the integral into two separate integrals. $\newline$ The integral of a sum is the sum of the integrals, so we can write: $\newline$ $\int(7e^{0.5x} - 5)\,dx = \int 7e^{0.5x}\,dx - \int 5\,dx$ from $0$ to $6$ .
Evaluate first integral: Evaluate the first integral $\int_{0}^{6} 7e^{(0.5x)}dx$ . The antiderivative of $e^{(0.5x)}$ is $(2e^{(0.5x)})$ , because when we differentiate $(2e^{(0.5x)})$ , we get $e^{(0.5x)}$ times the derivative of $(0.5x)$ which is $0.5$ , and $2 \times 0.5 = 1$ , giving us back $e^{(0.5x)}$ . Therefore, the antiderivative of $7e^{(0.5x)}$ is $e^{(0.5x)}$ $0$ .
Evaluate second integral: Evaluate the second integral $\int_{0}^{6} 5\,dx$ . The antiderivative of a constant $5$ is $5x$ . So the integral of $5$ from $0$ to $6$ is $5x$ evaluated from $0$ to $6$ .
Combine and evaluate: Combine the antiderivatives and evaluate from $0$ to $6$ . We have the antiderivatives $14e^{(0.5x)}$ and $5x$ . Now we need to evaluate these from $0$ to $6$ : $(14e^{(0.5x)} - 5x) |$ from $0$ to $6$ = $(14e^{(0.5\cdot6)} - 5\cdot6) - (14e^{(0.5\cdot0)} - 5\cdot0)$ .
Perform final evaluation: Perform the evaluation using the bounds $0$ and $6$ . First, we evaluate at the upper bound $x = 6$ : $14e^{(0.5\times6)} - 5\times6 = 14e^3 - 30$ . Next, we evaluate at the lower bound $x = 0$ : $14e^{(0.5\times0)} - 5\times0 = 14e^0 - 0 = 14\times1 - 0 = 14$ . Now, subtract the lower bound evaluation from the upper bound evaluation: $(14e^3 - 30) - (14) = 14e^3 - 30 - 14 = 14e^3 - 44$ .