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∫09x9−xdx\int_{0}^{9} \frac{x}{\sqrt{9-x}} d x \quad Using substitution method where $u...
Jan 27, 2024
∫09x9−xdx\int_{0}^{9} \frac{x}{\sqrt{9-x}} d x \quad Using substitution method where u2=9−xu^{2}=9-x
Solution by Steps
step 1
To solve the integral ∫09x9−xdx\int_{0}^{9} \frac{x}{\sqrt{9-x}} dx using the substitution method, let u2=9−xu^2 = 9 - x
step 2
Differentiate both sides with respect to xx to find dudx\frac{du}{dx}. This gives 2ududx=−12u \frac{du}{dx} = -1, so dudx=−12u\frac{du}{dx} = -\frac{1}{2u}
step 3
Solve for dxdx to substitute in the integral: dx=−2ududx = -2u du
step 4
Substitute u2=9−xu^2 = 9 - x into xx to get x=9−u2x = 9 - u^2
step 5
Substitute xx and dxdx into the integral: ∫099−u29−(9−u2)(−2u)du\int_{0}^{9} \frac{9 - u^2}{\sqrt{9-(9-u^2)}} (-2u) du
step 6
Simplify the integral: ∫099−u2u(−2u)du=∫09−2(9−u2)du\int_{0}^{9} \frac{9 - u^2}{u} (-2u) du = \int_{0}^{9} -2(9 - u^2) du
step 7
Integrate with respect to uu: −2∫09(9−u2)du=−2[9u−u33]09-2 \int_{0}^{9} (9 - u^2) du = -2 \left[ 9u - \frac{u^3}{3} \right]_{0}^{9}
step 8
Evaluate the integral from the bounds u=0u = 0 to u=3u = 3 (since u2=9−xu^2 = 9 - x and xx ranges from 0 to 9, uu ranges from 3 to 0): −2[9(3)−333]−(−2[9(0)−033])-2 \left[ 9(3) - \frac{3^3}{3} \right] - (-2 \left[ 9(0) - \frac{0^3}{3} \right])
step 9
Simplify the result: −2[27−9]−(−2[0−0])=−2[18]=−36-2 \left[ 27 - 9 \right] - (-2 \left[ 0 - 0 \right]) = -2 \left[ 18 \right] = -36
Answer
−36-36
Key Concept
Integration using substitution
Explanation
The integral of a function can be simplified by using a substitution that transforms the integral into a more manageable form. After substituting and integrating, the original variable is substituted back in to find the definite integral.
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