Evaluate int_(0)^(30)(2e^(-0.2 x)-1)dx and express the answer in simplest form. Answer:

Break down the integral: Break down the integral into two separate integrals. int_(0)^(30)(2e^(-0.2x) - 1)dx = int_(0)^(30)2e^(-0.2x)dx - int_(0)^(30)1dxEvaluate first integral: Evaluate the first integral int_(0)^(30)2e^(-0.2x)dx. Let u = -0.2x, then du = -0.2dx, which implies dx = -5du. The limits of integration change from x = 0 to x = 30 to u = 0 to u = -6. The integral becomes -10 * int_(0)^(-6)e^u du.Evaluate integral of e^u: Evaluate the integral of e^u. int e^u du = e^u + C So, -10 * int_(0)^(-6)e^u du = -10 * (e^(-6) - e^(0))Evaluate second integral: Evaluate the second integral int_(0)^(30)1dx. This is a simple integral, which results in x evaluated from 0 to 30. int_(0)^(30)1dx = 30 - 0 = 30Combine results: Combine the results from Step 3 and Step 4. -10 * (e^(-6) - e^(0)) - 30 Simplify the expression. -10 * (e^(-6) - 1) - 30Calculate exact value: Calculate the exact value of the expression. -10 * (1/e^(6) - 1) - 30 = -10 * (1/e^(6) - 1) - 30 = -10/e^(6) + 10 - 30 = -10/e^(6) - 20

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Evaluate
int_(0)^(30)(2e^(-0.2 x)-1)dx and express the answer in simplest form.
Answer:

Evaluate $\int_{0}^{30}\left(2 e^{-0.2 x}-1\right) d x$ 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}^{30}\left(2 e^{-0.2 x}-1\right) d x

and express the answer in simplest form.

\newline

Answer:

Break down the integral: Break down the integral into two separate integrals. \int_{$0$}^{$30$}(\(2e^{ $-0$ . $2$ x} - $1$ )\,dx = \int_{ $0$ }^{ $30$ } $2$ e^{ $-0$ . $2$ x}\,dx - \int_{ $0$ }^{ $30$ } $1$ \,dx
Evaluate first integral: Evaluate the first integral $\int_{0}^{30} 2e^{-0.2x}dx$ . Let $u = -0.2x$ , then $du = -0.2dx$ , which implies $dx = -5du$ . The limits of integration change from $x = 0$ to $x = 30$ to $u = 0$ to $u = -6$ . The integral becomes $-10 \times \int_{0}^{-6}e^{u} du$ .
Evaluate integral of $e^u$ : Evaluate the integral of $e^u$ .
$\int e^u du = e^u + C$
So, $-10 \times \int_{0}^{-6}e^u du = -10 \times (e^{-6} - e^{0})$
Evaluate second integral: Evaluate the second integral $\int_{0}^{30} 1 \, dx$ . This is a simple integral, which results in $x$ evaluated from $0$ to $30$ . $\int_{0}^{30} 1 \, dx = 30 - 0 = 30$
Combine results: Combine the results from Step $3$ and Step $4$ . $\newline$ $-10 \times (e^{-6} - e^{0}) - 30$ $\newline$ Simplify the expression. $\newline$ $-10 \times (e^{-6} - 1) - 30$
Calculate exact value: Calculate the exact value of the expression. $\newline$ $-10 \times (1/e^{6} - 1) - 30$ $\newline$ = $-10 \times (1/e^{6} - 1) - 30$ $\newline$ = $-10/e^{6} + 10 - 30$ $\newline$ = $-10/e^{6} - 20$