Evaluate int_(0)^(16)(5e^(-0.25 x)-3)dx and express the answer in simplest form. Answer:

Break down into two integrals: Break down the integral into two separate integrals. int_(0)^(16)(5e^(-0.25x) - 3)dx = int_(0)^(16)5e^(-0.25x)dx - int_(0)^(16)3dxEvaluate first integral: Evaluate the first integral int_(0)^(16)5e^(-0.25x)dx. Let u = -0.25x, then du = -0.25dx, which implies dx = -4du. The limits of integration change from x = 0 to x = 16 to u = 0 to u = -4. The integral becomes -20 * int_(0)^(-4)e^u du.Evaluate integral of e^u: Evaluate the integral of e^u. int e^u du = e^u + C So, -20 * int_(0)^(-4)e^u du = -20 * (e^(-4) - e^(0))Evaluate second integral: Evaluate the second integral int_(0)^(16)3dx. This is a simple integral, and the result is 3x evaluated from 0 to 16. So, int_(0)^(16)3dx = 3 * 16 - 3 * 0 = 48.Combine results: Combine the results from Step 3 and Step 4. -20 * (e^(-4) - e^(0)) + 48 = -20 * (e^(-4) - 1) + 48Simplify expression: Simplify the expression. -20 * (e^(-4) - 1) + 48 = -20 * (1/e^(4) - 1) + 48 = -20/e^(4) + 20 + 48 = 68 - 20/e^(4)

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Evaluate
int_(0)^(16)(5e^(-0.25 x)-3)dx and express the answer in simplest form.
Answer:

Evaluate $\int_{0}^{16}\left(5 e^{-0.25 x}-3\right) d x$ 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}^{16}\left(5 e^{-0.25 x}-3\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. $\int_{0}^{16}(5e^{-0.25x} - 3)dx = \int_{0}^{16}5e^{-0.25x}dx - \int_{0}^{16}3dx$
Evaluate first integral: Evaluate the first integral $\int_{0}^{16}5e^{-0.25x}\,dx$ . Let $u = -0.25x$ , then $du = -0.25dx$ , which implies $dx = -4du$ . The limits of integration change from $x = 0$ to $x = 16$ to $u = 0$ to $u = -4$ . The integral becomes $-20 \times \int_{0}^{-4}e^{u} \,du$ .
Evaluate integral of $e^u$ : Evaluate the integral of $e^u$ .
$\int e^u \, du = e^u + C$
So, $-20 \times \int_{0}^{-4}e^u \, du = -20 \times (e^{-4} - e^{0})$
Evaluate second integral: Evaluate the second integral $\int_{0}^{16} 3 \, dx$ . This is a simple integral, and the result is $3x$ evaluated from $0$ to $16$ . So, $\int_{0}^{16} 3 \, dx = 3 \times 16 - 3 \times 0 = 48$ .
Combine results: Combine the results from Step $3$ and Step $4$ . $\newline$ $-20 \times (e^{-4} - e^{0}) + 48 = -20 \times (e^{-4} - 1) + 48$
Simplify expression: Simplify the expression. $\newline$ $-20 * (e^{-4} - 1) + 48 = -20 * (\frac{1}{e^{4}} - 1) + 48$ $\newline$ $= \frac{-20}{e^{4}} + 20 + 48$ $\newline$ $= 68 - \frac{20}{e^{4}}$