Question

Evaluate the definite integral. Use a graphing utility to verify your result. $$ \int_{1}^{2}(x-1) \sqrt{2-x} d x $$

Short answer

The evaluation of the definite integral \(\int_{1}^{2}(x-1) \sqrt{2-x} d x\) is \(\frac{22}{15}\). The graphical representation of the function confirms this result.

Step-by-step solution

Apply the integration by parts

Integration by parts is a method based on the rule of product differentiation. In this case, let \( u = x - 1 \) and \( dv = \sqrt{2 - x} dx \). Consequently, the necessary derivatives and integrals are that \( du = dx \) and \( v = - \frac{2}{3} (2 - x)^{3/2} \). Now use the integration by parts formula: \( \int u dv = uv - \int v du \)

Apply the formula of integration by parts

Apply the formula to the integral: \( \int_{1}^{2}(x-1) \sqrt{2-x} d x = uv |_{1}^{2} - \int_{1}^{2} v du \). Substitute the values of u, v, du to get: \( =[- (x-1) \cdot \frac{2}{3}(2-x)^{3/2} ]_{1}^{2} - \int_{1}^{2} -\frac{2}{3}(2-x)^{3/2} dx \)

Simplify the integral

Simplifying the integral formula now becomes a simple integral: \( = [2 - \frac{4}{3}] + [\frac{2}{3} \int_{1}^{2} (2-x)^{3/2} dx] \). You can now move with the calculation of the simple integral.

Substitute to solve the integral

Let \( t = 2 - x \). Then \( dt = -dx \). Substitute these values into the integral and solve it: \( = 2 - \frac{4}{3} + \frac{2}{3} \int_{0}^{1} t^{3/2} (-dt) \) After putting everything together and solving, we get \( = 2 - \frac{4}{3} - \frac{2}{3} ( \frac{2}{5} t^{5/2}|_{0}^{1} ) \).

Final evaluation of the integral

Perform the final calculations to get the definite integral value: \( = 2 - \frac{4}{3} - \frac{4}{15} = \frac{22}{15} \)

Verify the result with a graphing utility

Plot the function \(f(x) = (x-1) \sqrt{2-x}\) for the interval \([1, 2]\) and calculate the area under the curve to confirm the result. If done correctly, the area under the curve should agree with the solution to the integral which is \( \frac{22}{15} \).