Evaluate the indefinite integral.∫(x−ex)dx\int_{}^{}\left(\sqrt{x}-e^{x}\right)dx∫(x−ex)dx

2 3 x 3 2 − e x + C \frac23x^{\frac32}-e^{x}+C 3 2 x 2 3 − e x + C

Evaluate the indefinite integral. ∫ ( x − e x ) d x \int_{}^{}\left(\sqrt{x}-e^{x}\right)dx ∫ ( x <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> − e x ) d x

we're asked to evaluate the indefinite integral of the square root of x minus e to the x d x. Now looking at this integral, I know that the square root of x is the same thing as x to the power of one half, so I can go ahead and use the general power rule to find the antiderivative of this first term. So using my general power rule here, that makes this two thirds times x to the power of three halves. Then subtracting off this next anti the antiderivative of my next term here e to the x. I know that is also still e to the x, so subtracting e to the x and then of course adding my constant of integration c and this gives me my final answer. Two thirds x to the power of three halves minus e to the power of x plus c. Let us know if you have any questions and let's keep going.

Evaluate the indefinite integral. ∫ ( x − e x ) d x \int_{}^{}\left(\sqrt{x}-e^{x}\right)dx ∫ ( x <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> − e x ) d x

2 3 x 3 2 − e x + C \frac23x^{\frac32}-e^{x}+C 3 2 x 2 3 − e x + C

2 3 x 2 3 − e x + C \frac23x^{\frac23}-e^{x}+C 3 2 x 3 2 − e x + C

2 3 x 3 2 − x e x + C \frac23x^{\frac32}-xe^{x}+C 3 2 x 2 3 − x e x + C

2 3 x 2 3 − 1 x e x + C \frac23x^{\frac23}-\frac{1}{x}e^{x}+C 3 2 x 3 2 − x 1 e x + C

10. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions

Business Calculus

10. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions

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Multiple Choice

Evaluate the indefinite integral.
$\int_{}^{}\left(\sqrt{x}-e^{x}\right)dx$

$\frac23x^{\frac23}-e^{x}+C$

$\frac23x^{\frac32}-xe^{x}+C$

$\frac23x^{\frac32}-e^{x}+C$

$\frac23x^{\frac23}-\frac{1}{x}e^{x}+C$

Verified step by step guidance

Step 1: Break the integral into two separate terms for easier evaluation. The given integral is ∫(√x - e^x) dx, which can be rewritten as ∫√x dx - ∫e^x dx.

Step 2: For the first term, ∫√x dx, rewrite √x as x^(1/2). Use the power rule for integration, which states that ∫x^n dx = (x^(n+1))/(n+1) + C, where n ≠ -1.

Step 3: Apply the power rule to x^(1/2). Add 1 to the exponent (1/2 + 1 = 3/2) and divide by the new exponent (3/2). This gives (2/3)x^(3/2).

Step 4: For the second term, ∫e^x dx, recall that the integral of e^x is simply e^x. So, the result of this term is -e^x (note the negative sign from the original integral).

Step 5: Combine the results of both terms. The final expression for the indefinite integral is (2/3)x^(3/2) - e^x + C, where C is the constant of integration.

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