Math / Calculus

Question $\int_{4}^{5} x \sqrt{x^{2}-16} d x$

Studdy Solution

STEP 1

1. The integral $\int_{4}^{5} x \sqrt{x^{2}-16} \, dx$ is a definite integral.
2. We will use substitution to simplify the integral.
3. The limits of integration will need to be adjusted according to the substitution.

STEP 2

1. Choose an appropriate substitution to simplify the integrand.
2. Change the limits of integration according to the substitution.
3. Perform the integration.
4. Evaluate the definite integral.

STEP 3

Choose a substitution. Let $u = x^2 - 16$ . Then, differentiate $u$ with respect to $x$ :
$\frac{du}{dx} = 2x$ $du = 2x \, dx$
Solve for $x \, dx$ :
$x \, dx = \frac{1}{2} du$

STEP 4

Change the limits of integration. When $x = 4$ , calculate $u$ :
$u = 4^2 - 16 = 0$
When $x = 5$ , calculate $u$ :
$u = 5^2 - 16 = 9$
Thus, the new limits of integration are from $u = 0$ to $u = 9$ .

STEP 5

Substitute into the integral:
$\int_{4}^{5} x \sqrt{x^{2}-16} \, dx = \int_{0}^{9} \sqrt{u} \cdot \frac{1}{2} \, du$
Simplify the integral:
$\frac{1}{2} \int_{0}^{9} u^{1/2} \, du$

STEP 6

Perform the integration:
$\frac{1}{2} \int u^{1/2} \, du = \frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right]_{0}^{9}$
$= \frac{1}{2} \cdot \frac{2}{3} \left[ u^{3/2} \right]_{0}^{9}$
$= \frac{1}{3} \left[ u^{3/2} \right]_{0}^{9}$

STEP 7

Evaluate the definite integral:
$= \frac{1}{3} \left[ 9^{3/2} - 0^{3/2} \right]$
Calculate $9^{3/2}$ :
$9^{3/2} = (9^{1/2})^3 = 3^3 = 27$
So, the integral evaluates to:
$= \frac{1}{3} \times 27 = 9$
The value of the definite integral is:
$\boxed{9}$

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