Questions: Evaluate the definite integral of the algebraic function. Type the exact answer. ∫ from 1 to 9 of (u-7)/√u du

Solution Steps

To evaluate the definite integral of the given algebraic function, we can first simplify the integrand by splitting it into two separate integrals. Then, we can use the power rule for integration to find the antiderivative of each term. Finally, we will evaluate the antiderivatives at the given bounds and subtract to find the exact value of the definite integral.

We start with the definite integral: \[ \int_{1}^{9} \frac{u-7}{\sqrt{u}} \, du \] We can split this integral into two parts: \[ \int_{1}^{9} \frac{u}{\sqrt{u}} \, du - \int_{1}^{9} \frac{7}{\sqrt{u}} \, du \]

The first term simplifies as follows: \[ \frac{u}{\sqrt{u}} = u^{1/2} \] The second term can be rewritten as: \[ \frac{7}{\sqrt{u}} = 7u^{-1/2} \] Thus, we can express the integral as: \[ \int_{1}^{9} u^{1/2} \, du - 7 \int_{1}^{9} u^{-1/2} \, du \]

The antiderivative of \( u^{1/2} \) is: \[ \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2} \] The antiderivative of \( 7u^{-1/2} \) is: \[ 7 \cdot 2u^{1/2} = 14u^{1/2} \] Now we evaluate both antiderivatives from 1 to 9.

Evaluating the first integral: \[ \left[ \frac{2}{3} u^{3/2} \right]_{1}^{9} = \frac{2}{3} (9^{3/2} - 1^{3/2}) = \frac{2}{3} (27 - 1) = \frac{2}{3} \cdot 26 = \frac{52}{3} \] Evaluating the second integral: \[ \left[ 14u^{1/2} \right]_{1}^{9} = 14 (9^{1/2} - 1^{1/2}) = 14 (3 - 1) = 14 \cdot 2 = 28 \]

Now we combine the results of the two integrals: \[ \frac{52}{3} - 28 = \frac{52}{3} - \frac{84}{3} = \frac{52 - 84}{3} = \frac{-32}{3} \]

Final Answer