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int_(1)^(4)(3x+sqrtx)dx=

$\int_{1}^{4}(3 x+\sqrt{x}) d x=$

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Evaluate definite integrals using the power rule

Full solution

Q.

\int_{1}^{4}(3 x+\sqrt{x}) d x=

Break into two integrals: We need to integrate the function $3x + \sqrt{x}$ from $1$ to $4$ . Break it into two separate integrals: $\newline$ $\int_1^4 (3x + \sqrt{x}) \, dx = \int_1^4 3x \, dx + \int_1^4 \sqrt{x} \, dx$
Calculate $3$ x integral: First, calculate $\int_1^4 3x \, dx$ : $\newline$ $\int 3x \, dx = \frac{3}{2}x^2 + C$ $\newline$ Evaluating from $1$ to $4$ : $\newline$ $\left[\frac{3}{2}x^2\right]_1^4 = \frac{3}{2}(4^2) - \frac{3}{2}(1^2) = \frac{3}{2}(16) - \frac{3}{2}(1) = 24 - 1.5 = 22.5$
Calculate sqrt(x) integral: Next, calculate $\int_1^4 \sqrt{x} \, dx$ : $\newline$ $\int \sqrt{x} \, dx = \frac{2}{3}x^{3/2} + C$ $\newline$ Evaluating from $1$ to $4$ : $\newline$ $\left[\frac{2}{3}x^{3/2}\right]_1^4 = \frac{2}{3}(4^{3/2}) - \frac{2}{3}(1^{3/2}) = \frac{2}{3}(8) - \frac{2}{3}(1) = \frac{16}{3} - \frac{2}{3} = \frac{14}{3}$
Add results: Add the results of the two integrals: $\newline$ $22.5 + \frac{14}{3} = 22.5 + 4.67 = 27.17$

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