Problem 20 - Calculus Early Transcendentals

Question

Evaluate the following integrals as they are written. $$\int_{0}^{3} \int_{x^{2}}^{x+6}(x-1) d y d x$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral $$\int_{0}^{3} \int_{x^{2}}^{x+6}(x-1) d y d x$$ Answer: The result of the definite integral is: $$\int_{0}^{3} \int_{x^{2}}^{x+6}(x-1) d y d x= -\frac{23}{4}$$

Step 1: Perform Inner Integral (dy)

First, focus on the inner integral, which is with respect to y. Since the integrand \((x-1)\) does not contain any y terms, we can simply multiply it by the difference of the y limits:$$\int_{x^2}^{x+6} (x-1)dy = (x-1) \int_{x^2}^{x+6} dy = (x-1) [y]_{x^2}^{x+6}$$Now, evaluate the limits of integration for y:$$(x-1) [y]_{x^2}^{x+6} = (x-1)((x+6)-x^2)$$Next, simplify the expression:$$=(x-1)(6-x^2+x)$$Now we have to perform the outer integral.

Step 2: Perform Outer Integral (dx)

Now, we will integrate the obtained result in Step 1, with the limits for x from 0 to 3:$$\int_{0}^{3} (x-1)(6-x^2+x)dx$$In order to perform this integral, we will first, distribute the (x-1) term to (6-x^2+x):$$\int_{0}^{3} (-x^3 + x^2+x-6)dx$$Now, integrate each term with respect to x, remembering to put the limits of integration (0 to 3):$$[-\frac{x^4}{4} + \frac{x^3}{3}+\frac{x^2}{2}-6x]_{0}^{3}$$Finally, evaluate the limits:

Step 3: Evaluate the definite integral

Plug the upper limit (3) into the integrated function, and then plug the lower limit (0) into the function. Subtract the lower value from the upper value:$$[-\frac{(3)^4}{4} + \frac{(3)^3}{3}+\frac{(3)^2}{2}-6(3)]-[-\frac{(0)^4}{4} + \frac{(0)^3}{3}+\frac{(0)^2}{2}-6(0)]$$Now, simplify the expression to find the result:$$=>-\frac{81}{4} + 9+\frac{9}{2}-18-(0)$$$$=>-\frac{81}{4} + 9+\frac{9}{2}-18$$$$=>(-\frac{81}{4} + \frac{58}{4})$$$$=>\frac{-23}{4}$$Therefore, the result of the integral is: $$\int_{0}^{3} \int_{x^{2}}^{x+6}(x-1) d y d x= -\frac{23}{4}$$