Evaluate int_(0)^(15)(9e^(-0.2 x)-1)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 9e^(-0.2x) - 1 from 0 to 15. This can be written as: ∫ from 0 to 15 of (9e^(-0.2x) - 1) dxBreak into two: Break the integral into two separate integrals. The integral of a sum is the sum of the integrals, so we can write: ∫ from 0 to 15 of 9e^(-0.2x) dx - ∫ from 0 to 15 of 1 dxEvaluate first integral: Evaluate the first integral. The integral of 9e^(-0.2x) with respect to x is: ∫ 9e^(-0.2x) dx = -45e^(-0.2x) + C We will evaluate this from 0 to 15 in the next steps.Evaluate second integral: Evaluate the second integral. The integral of 1 with respect to x is: ∫ 1 dx = x + C We will evaluate this from 0 to 15 in the next steps.Apply Fundamental Theorem: Apply the Fundamental Theorem of Calculus to the first integral. We need to evaluate -45e^(-0.2x) from 0 to 15: -45e^(-0.2 * 15) + 45e^(-0.2 * 0)Calculate first integral values: Calculate the values for the first integral. -45e^(-3) + 45e^(0) Since e^(0) = 1, this simplifies to: -45e^(-3) + 45Apply Fundamental Theorem: Apply the Fundamental Theorem of Calculus to the second integral. We need to evaluate x from 0 to 15: 15 - 0 This simplifies to 15.Combine results: Combine the results from the two integrals. Now we combine the results from steps 6 and 7: (-45e^(-3) + 45) - 15Simplify expression: Simplify the expression. Simplify the expression to get the final answer: -45e^(-3) + 45 - 15 -45e^(-3) + 30

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Evaluate
int_(0)^(15)(9e^(-0.2 x)-1)dx and express the answer in simplest form.
Answer:

Evaluate $\int_{0}^{15}\left(9 e^{-0.2 x}-1\right) d x$ and express the answer in simplest form. $\newline$ Answer:

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Calculus

Evaluate definite integrals using the chain rule

Full solution

Q. Evaluate

\int_{0}^{15}\left(9 e^{-0.2 x}-1\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 $9e^{-0.2x} - 1$ from $0$ to $15$ . This can be written as: $\newline$ $\int_{0}^{15} (9e^{-0.2x} - 1) \, dx$
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: $\newline$ $\int_{0}^{15} 9e^{-0.2x} \, dx - \int_{0}^{15} 1 \, dx$
Evaluate first integral: Evaluate the first integral. $\newline$ The integral of $9e^{-0.2x}$ with respect to $x$ is: $\newline$ $\int 9e^{-0.2x} dx = -45e^{-0.2x} + C$ $\newline$ We will evaluate this from $0$ to $15$ in the next steps.
Evaluate second integral: Evaluate the second integral. $\newline$ The integral of $1$ with respect to $x$ is: $\newline$ $\int 1 \, dx = x + C$ $\newline$ We will evaluate this from $0$ to $15$ in the next steps.
Apply Fundamental Theorem: Apply the Fundamental Theorem of Calculus to the first integral. $\newline$ We need to evaluate $-45e^{(-0.2x)}$ from $0$ to $15$ : $\newline$ $-45e^{(-0.2 \times 15)} + 45e^{(-0.2 \times 0)}$
Calculate first integral values: Calculate the values for the first integral. $\newline$ $-45e^{-3} + 45e^{0}$ $\newline$ Since $e^{0} = 1$ , this simplifies to: $\newline$ $-45e^{-3} + 45$
Apply Fundamental Theorem: Apply the Fundamental Theorem of Calculus to the second integral. $\newline$ We need to evaluate $x$ from $0$ to $15$ : $\newline$ $15 - 0$ $\newline$ This simplifies to $15$ .
Combine results: Combine the results from the two integrals. $\newline$ Now we combine the results from steps $6$ and $7$ : $\newline$ $(-45e^{-3} + 45) - 15$
Simplify expression: Simplify the expression. $\newline$ Simplify the expression to get the final answer: $\newline$ $-45e^{-3} + 45 - 15$ $\newline$ $-45e^{-3} + 30$