Evaluate int_(0)^(2)(6e^(-0.5 x)-4x)dx and express the answer in simplest form. Answer:

Break down into two integrals: Break down the integral into two separate integrals. int_(0)^(2)(6e^(-0.5x) - 4x)dx = int_(0)^(2)6e^(-0.5x)dx - int_(0)^(2)4xdxEvaluate integral of 6e^(-0.5x): Evaluate the first integral int_(0)^(2)6e^(-0.5x)dx. Let u = -0.5x, then du = -0.5dx, which implies dx = -2du. When x = 0, u = 0, and when x = 2, u = -1. The integral becomes -12 int_(0)^(-1)e^u du.Calculate integral of e^u: Calculate the integral of e^u. The integral of e^u with respect to u is e^u. So, -12 int_(0)^(-1)e^u du = -12[e^u]_(0)^(-1) = -12(e^(-1) - e^(0)) = -12(1/e - 1).Evaluate integral of 4x: Evaluate the second integral int_(0)^(2)4xdx. The integral of x with respect to x is (1/2)x^2. So, int_(0)^(2)4xdx = 4[(1/2)x^2]_(0)^(2) = 4(1/2)(2)^2 - 4(1/2)(0)^2 = 4(2) - 0 = 8.Combine results from Step 3 and Step 4: Combine the results from Step 3 and Step 4. The result of the integral from 0 to 2 of the function 6e^(-0.5x) - 4x is -12(1/e - 1) + 8.Simplify the expression: Simplify the expression. -12(1/e - 1) + 8 = -12/e + 12 + 8 = -12/e + 20.

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Evaluate
int_(0)^(2)(6e^(-0.5 x)-4x)dx and express the answer in simplest form.
Answer:

Evaluate $\int_{0}^{2}\left(6 e^{-0.5 x}-4 x\right) d x$ and express the answer in simplest form. $\newline$ Answer:

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Calculus

Evaluate definite integrals using the power rule

Full solution

Q. Evaluate

\int_{0}^{2}\left(6 e^{-0.5 x}-4 x\right) d x

and express the answer in simplest form.

\newline

Answer:

Break down into two integrals: Break down the integral into two separate integrals. $\int_{0}^{2}(6e^{-0.5x} - 4x)dx = \int_{0}^{2}6e^{-0.5x}dx - \int_{0}^{2}4xdx$
Evaluate integral of $6e^{-0.5x}$ : Evaluate the first integral $\int_{0}^{2}6e^{-0.5x}\,dx$ . Let $u = -0.5x$ , then $du = -0.5dx$ , which implies $dx = -2du$ . When $x = 0$ , $u = 0$ , and when $x = 2$ , $u = -1$ . The integral becomes $-12 \int_{0}^{-1}e^{u} \,du$ .
Calculate integral of $e^u$ : Calculate the integral of $e^u$ . The integral of $e^u$ with respect to $u$ is $e^u$ . So, $-12 \int_{0}^{-1}e^u du = -12[e^u]_{0}^{-1} = -12(e^{-1} - e^{0}) = -12(\frac{1}{e} - 1)$ .
Evaluate integral of $4x$ : Evaluate the second integral $\int_{0}^{2}4x\,dx$ . The integral of $x$ with respect to $x$ is $(1/2)x^2$ . So, $\int_{0}^{2}4x\,dx = 4[(1/2)x^2]_{0}^{2} = 4(1/2)(2)^2 - 4(1/2)(0)^2 = 4(2) - 0 = 8$ .
Combine results from Step $3$ and Step $4$ : Combine the results from Step $3$ and Step $4$ . The result of the integral from $0$ to $2$ of the function $6e^{-0.5x} - 4x$ is $-12(1/e - 1) + 8$ .
Simplify the expression: Simplify the expression. $\newline$ $-12(\frac{1}{e} - 1) + 8 = \frac{-12}{e} + 12 + 8 = \frac{-12}{e} + 20$ .