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Definite Integration: 16x 16x

JEE Maths question with a full step-by-step solution.

Question
∫48(16x−64+x+x−16x−64)dx=\int_4^8\left(\sqrt{\sqrt{16x-64}+x}+\sqrt{x-\sqrt{16x-64}}\right)dx =
A2∫46(16x−64+x+x−16x−64)dx2\displaystyle\int_4^6\left(\sqrt{\sqrt{16x-64}+x}+\sqrt{x-\sqrt{16x-64}}\right)dxcorrect
B2∫810(16x−64+x+x−16x−64)dx2\displaystyle\int_8^{10}\left(\sqrt{\sqrt{16x-64}+x}+\sqrt{x-\sqrt{16x-64}}\right)dx
C2∫79(16x−64+x+x−16x−64)dx2\displaystyle\int_7^9\left(\sqrt{\sqrt{16x-64}+x}+\sqrt{x-\sqrt{16x-64}}\right)dx
D1616correct
Solution
Step 1:
16x−64=16(x−4)=4x−4(x≥4)\sqrt{16x-64} = \sqrt{16\left(x-4\right)} = 4\sqrt{x-4}\qquad\left(x \ge 4\right)
 Step 2: Writing u=x−4u = \sqrt{x-4}, so that x=u2+4x = u^2+4,
x+4x−4=u2+4u+4=(u+2)2x+4\sqrt{x-4} = u^2+4u+4 = \left(u+2\right)^2
x−4x−4=u2−4u+4=(u−2)2x-4\sqrt{x-4} = u^2-4u+4 = \left(u-2\right)^2
 Step 3: u=x−4≥0u = \sqrt{x-4} \ge 0, so
f(x)=∣u+2∣+∣u−2∣=(u+2)+∣u−2∣f\left(x\right) = \left|u+2\right|+\left|u-2\right| = \left(u+2\right)+\left|u-2\right|
 Step 4: u≤2u \le 2 means x≤8x \le 8, so
f(x)={(u+2)+(2−u)=4,4≤x≤8,(u+2)+(u−2)=2u=2x−4,x≥8.f\left(x\right) = \begin{cases} \left(u+2\right)+\left(2-u\right) = 4 , & 4 \le x \le 8 ,\\[1mm] \left(u+2\right)+\left(u-2\right) = 2u = 2\sqrt{x-4} , & x \ge 8 . \end{cases}
 So ff is constantly 44 on the whole interval of integration. Step 5:
∫48f(x)dx=∫484 dx=4×4=16\int_4^8 f\left(x\right)dx = \int_4^8 4\,dx = 4\times4 = 16
 (4) holds. Step 6: On [4,6]\left[4,6\right] we still have x≤8x \le 8, so f=4f = 4 throughout:
2∫464 dx=2×4×2=162\int_4^6 4\,dx = 2\times4\times2 = 16
 (1) holds. Step 7: On [8,10]\left[8,10\right] we are past the corner, so f=2x−4f = 2\sqrt{x-4}:
2∫8102x−4 dx=4⋅23[(x−4)3/2]810=83(66−8)=83(14.6969−8)=17.8585≠162\int_8^{10}2\sqrt{x-4}\,dx = 4\cdot\frac23\left[\left(x-4\right)^{3/2}\right]_8^{10} = \frac83\left(6\sqrt6-8\right) = \frac83\left(14.6969-8\right) = 17.8585 \ne 16
 (2) fails. Step 8: [7,9]\left[7,9\right] straddles the corner at x=8x = 8:
∫784 dx=4,∫892x−4 dx=43(55−8)=4.2404\int_7^8 4\,dx = 4 ,\qquad \int_8^9 2\sqrt{x-4}\,dx = \frac43\left(5\sqrt5-8\right) = 4.2404
2(4+4.2404)=16.4809≠162\left(4+4.2404\right) = 16.4809 \ne 16
 (3) fails. Answer: (1) and (4)
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