Evaluate int_(0)^(5)(3e^(-0.2 x)-2x)dx and express the answer in simplest form. Answer:

Break down the integral: Break down the integral into two separate integrals. int_(0)^(5)(3e^(-0.2x) - 2x)dx = int_(0)^(5)3e^(-0.2x)dx - int_(0)^(5)2xdxEvaluate first integral: Evaluate the first integral int_(0)^(5)3e^(-0.2x)dx. Let u = -0.2x, then du = -0.2dx, which implies dx = -5du. The limits of integration change from x = 0 to u = 0 and from x = 5 to u = -1. The integral becomes -15 int_(0)^(-1)e^u du.Calculate integral of e^u: Calculate the integral of e^u. int e^u du = e^u + C So, -15 int_(0)^(-1)e^u du = -15(e^u)|_(0)^(-1) = -15(e^(-1) - e^(0)) = -15(1/e - 1)Evaluate second integral: Evaluate the second integral int_(0)^(5)2xdx. int x dx = (1/2)x^2 + C So, int_(0)^(5)2xdx = 2 * (1/2)x^2|_(0)^(5) = x^2|_(0)^(5) = 5^2 - 0^2 = 25Combine results: Combine the results from Step 3 and Step 4. int_(0)^(5)(3e^(-0.2x) - 2x)dx = -15(1/e - 1) + 25Simplify expression: Simplify the expression. -15(1/e - 1) + 25 = -15/e + 15 + 25 = 40 - 15/e

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Evaluate
int_(0)^(5)(3e^(-0.2 x)-2x)dx and express the answer in simplest form.
Answer:

Evaluate $\int_{0}^{5}\left(3 e^{-0.2 x}-2 x\right) d x$ 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}^{5}\left(3 e^{-0.2 x}-2 x\right) d x

and express the answer in simplest form.

\newline

Answer:

Break down the integral: Break down the integral into two separate integrals. $\int_{0}^{5}(3e^{-0.2x} - 2x)dx = \int_{0}^{5}3e^{-0.2x}dx - \int_{0}^{5}2xdx$
Evaluate first integral: Evaluate the first integral $\int_{0}^{5} 3e^{-0.2x} \, dx$ . Let $u = -0.2x$ , then $du = -0.2dx$ , which implies $dx = -5du$ . The limits of integration change from $x = 0$ to $u = 0$ and from $x = 5$ to $u = -1$ . The integral becomes $-15 \int_{0}^{-1} e^{u} \, du$ .
Calculate integral of $e^u$ : Calculate the integral of $e^u$ . $\int e^u \, du = e^u + C$ So, \(-15 \int_{ $0$ }^{ $-1$ }e^u \, du = $-15$ (e^u)|_{ $0$ }^{ $-1$ } = $-15$ (e^{ $-1$ } - e^{ $0$ }) = $-15$ (\frac{ $1$ }{e} - $1$ )\]
Evaluate second integral: Evaluate the second integral $\int_{0}^{5} 2x \, dx$ . $\int x \, dx = \frac{1}{2}x^2 + C$ So, $\int_{0}^{5}2x\,dx = 2 \cdot \left(\frac{1}{2}x^2\right)\bigg|_{0}^{5} = x^2\bigg|_{0}^{5} = 5^2 - 0^2 = 25$
Combine results: Combine the results from Step $3$ and Step $4$ . $\int_{0}^{5}(3e^{-0.2x} - 2x)\,dx = -15(\frac{1}{e} - 1) + 25$
Simplify expression: Simplify the expression. $\newline$ $-15(\frac{1}{e} - 1) + 25 = \frac{-15}{e} + 15 + 25 = 40 - \frac{15}{e}$