Evaluate int_(0)^(6)(10e^(-0.5 x)-2x)dx and express the answer in simplest form. Answer:

Break down integral: Break down the integral into two separate integrals. We have the integral of a sum of two functions, which can be separated into the sum of two integrals: ∫(10e^(-0.5x) - 2x)dx = ∫10e^(-0.5x)dx - ∫2xdxEvaluate first integral: Evaluate the first integral ∫10e^(-0.5x)dx. Let u = -0.5x, then du = -0.5dx, which implies dx = -2du. The limits of integration also change with the substitution. When x = 0, u = 0, and when x = 6, u = -3. The integral becomes: ∫10e^u * -2du = -20∫e^u du The integral of e^u with respect to u is e^u, so we have: -20e^u + C Substituting back for u, we get: -20e^(-0.5x) + CEvaluate second integral: Evaluate the second integral ∫2xdx. The integral of x with respect to x is (1/2)x^2, so we have: ∫2xdx = 2 * (1/2)x^2 = x^2Combine results: Combine the results from Step 2 and Step 3. The combined indefinite integral is: -20e^(-0.5x) + x^2 + CEvaluate definite integral: Evaluate the definite integral from 0 to 6. We need to calculate the value of the combined integral at the upper limit (x = 6) and subtract the value at the lower limit (x = 0). For the first part, -20e^(-0.5x), at x = 6 we have -20e^(-3), and at x = 0 we have -20e^(0) = -20. For the second part, x^2, at x = 6 we have 6^2 = 36, and at x = 0 we have 0^2 = 0. So the definite integral is: (-20e^(-3) + 36) - (-20 + 0)Simplify expression: Simplify the expression. Simplify the expression by combining like terms and calculating the value of e^(-3). (-20/e^3 + 36) - (-20) = -20/e^3 + 36 + 20 = 56 - 20/e^3

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Evaluate
int_(0)^(6)(10e^(-0.5 x)-2x)dx and express the answer in simplest form.
Answer:

Evaluate $\int_{0}^{6}\left(10 e^{-0.5 x}-2 x\right) d x$ and express the answer in simplest form. $\newline$ Answer:

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Calculus

Evaluate definite integrals using the chain rule

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Q. Evaluate

\int_{0}^{6}\left(10 e^{-0.5 x}-2 x\right) d x

and express the answer in simplest form.

\newline

Answer:

Break down integral: Break down the integral into two separate integrals. $\newline$ We have the integral of a sum of two functions, which can be separated into the sum of two integrals: $\newline$ \int(\(10e^{( $-0$ . $5$ x)} - $2$ x)\,dx = \int $10$ e^{( $-0$ . $5$ x)}\,dx - \int $2$ x\,dx
Evaluate first integral: Evaluate the first integral $\int_{0}^{10}e^{-0.5x}\,dx$ . Let $u = -0.5x$ , then $du = -0.5dx$ , which implies $dx = -2du$ . The limits of integration also change with the substitution. When $x = 0$ , $u = 0$ , and when $x = 6$ , $u = -3$ . The integral becomes: $\int_{0}^{10}e^{u} \cdot -2du = -20\int e^{u} \,du$ The integral of $e^{u}$ with respect to $u = -0.5x$ $0$ is $e^{u}$ , so we have: $u = -0.5x$ $2$ Substituting back for $u = -0.5x$ $0$ , we get: $u = -0.5x$ $4$
Evaluate second integral: Evaluate the second integral $\int 2x\,dx$ . The integral of $x$ with respect to $x$ is $(1/2)x^2$ , so we have: $\int 2x\,dx = 2 \times (1/2)x^2 = x^2$
Combine results: Combine the results from Step $2$ and Step $3$ . $\newline$ The combined indefinite integral is: $\newline$ $-20e^{(-0.5x)} + x^2 + C$
Evaluate definite integral: Evaluate the definite integral from $0$ to $6$ . We need to calculate the value of the combined integral at the upper limit ( $x = 6$ ) and subtract the value at the lower limit ( $x = 0$ ). For the first part, $-20e^{-0.5x}$ , at $x = 6$ we have $-20e^{-3}$ , and at $x = 0$ we have $-20e^{0} = -20$ . For the second part, $x^2$ , at $x = 6$ we have $6$ $1$ , and at $x = 0$ we have $6$ $3$ . So the definite integral is: $6$ $4$
Simplify expression: Simplify the expression. $\newline$ Simplify the expression by combining like terms and calculating the value of $e^{-3}$ . $\newline$ $(-20/e^3 + 36) - (-20)$ $\newline$ $= -20/e^3 + 36 + 20$ $\newline$ $= 56 - 20/e^3$