Evaluate the integral
We need to compute
I = ∫ 0 1 1 3 + x + 1 + x d x . I=\int_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}}\,dx. I = ∫ 0 1 3 + x + 1 + x 1 d x .
Rationalize the denominator
Use
1 3 + x + 1 + x = 3 + x − 1 + x ( 3 + x ) − ( 1 + x ) = 3 + x − 1 + x 2 . \frac{1}{\sqrt{3+x}+\sqrt{1+x}}
=\frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)}
=\frac{\sqrt{3+x}-\sqrt{1+x}}{2}. 3 + x + 1 + x 1 = ( 3 + x ) − ( 1 + x ) 3 + x − 1 + x = 2 3 + x − 1 + x . So,
I = 1 2 ∫ 0 1 ( 3 + x − 1 + x ) d x . I=\frac12\int_0^1 \left(\sqrt{3+x}-\sqrt{1+x}\right)dx. I = 2 1 ∫ 0 1 ( 3 + x − 1 + x ) d x .
Split the integral
I = 1 2 ( ∫ 0 1 3 + x d x − ∫ 0 1 1 + x d x ) . I=\frac12\left(\int_0^1 \sqrt{3+x}\,dx-\int_0^1 \sqrt{1+x}\,dx\right). I = 2 1 ( ∫ 0 1 3 + x d x − ∫ 0 1 1 + x d x ) . Now use
∫ x + k d x = 2 3 ( x + k ) 3 / 2 . \int \sqrt{x+k}\,dx=\frac{2}{3}(x+k)^{3/2}. ∫ x + k d x = 3 2 ( x + k ) 3/2 . Thus,
∫ 0 1 3 + x d x = 2 3 [ ( 3 + x ) 3 / 2 ] 0 1 = 2 3 ( 4 3 / 2 − 3 3 / 2 ) . \int_0^1 \sqrt{3+x}\,dx=\frac{2}{3}\left[(3+x)^{3/2}\right]_0^1
=\frac{2}{3}\left(4^{3/2}-3^{3/2}\right). ∫ 0 1 3 + x d x = 3 2 [ ( 3 + x ) 3/2 ] 0 1 = 3 2 ( 4 3/2 − 3 3/2 ) . Since
4 3 / 2 = 8 , 3 3 / 2 = 3 3 , 4^{3/2}=8,\qquad 3^{3/2}=3\sqrt3, 4 3/2 = 8 , 3 3/2 = 3 3 , we get
∫ 0 1 3 + x d x = 2 3 ( 8 − 3 3 ) = 16 3 − 2 3 . \int_0^1 \sqrt{3+x}\,dx=\frac{2}{3}(8-3\sqrt3)=\frac{16}{3}-2\sqrt3. ∫ 0 1 3 + x d x = 3 2 ( 8 − 3 3 ) = 3 16 − 2 3 . Also,
∫ 0 1 1 + x d x = 2 3 [ ( 1 + x ) 3 / 2 ] 0 1 = 2 3 ( 2 3 / 2 − 1 ) . \int_0^1 \sqrt{1+x}\,dx=\frac{2}{3}\left[(1+x)^{3/2}\right]_0^1
=\frac{2}{3}(2^{3/2}-1). ∫ 0 1 1 + x d x = 3 2 [ ( 1 + x ) 3/2 ] 0 1 = 3 2 ( 2 3/2 − 1 ) . Since
2 3 / 2 = 2 2 , 2^{3/2}=2\sqrt2, 2 3/2 = 2 2 , we get
∫ 0 1 1 + x d x = 2 3 ( 2 2 − 1 ) = 4 2 3 − 2 3 . \int_0^1 \sqrt{1+x}\,dx=\frac{2}{3}(2\sqrt2-1)=\frac{4\sqrt2}{3}-\frac{2}{3}. ∫ 0 1 1 + x d x = 3 2 ( 2 2 − 1 ) = 3 4 2 − 3 2 .
Substitute back
I = 1 2 ( 16 3 − 2 3 − ( 4 2 3 − 2 3 ) ) . I=\frac12\left(\frac{16}{3}-2\sqrt3-\left(\frac{4\sqrt2}{3}-\frac{2}{3}\right)\right). I = 2 1 ( 3 16 − 2 3 − ( 3 4 2 − 3 2 ) ) . Simplify inside:
16 3 + 2 3 = 6. \frac{16}{3}+\frac{2}{3}=6. 3 16 + 3 2 = 6. So,
I = 1 2 ( 6 − 4 2 3 − 2 3 ) = 3 − 2 2 3 − 3 . I=\frac12\left(6-\frac{4\sqrt2}{3}-2\sqrt3\right)
=3-\frac{2\sqrt2}{3}-\sqrt3. I = 2 1 ( 6 − 3 4 2 − 2 3 ) = 3 − 3 2 2 − 3 . Hence in the form
I = a + b 2 + c 3 , I=a+b\sqrt2+c\sqrt3, I = a + b 2 + c 3 , we have
a = 3 , b = − 2 3 , c = − 1. a=3,\qquad b=-\frac23,\qquad c=-1. a = 3 , b = − 3 2 , c = − 1.
Compute the required expression
2 a + 3 b − 4 c = 2 ( 3 ) + 3 ( − 2 3 ) − 4 ( − 1 ) . 2a+3b-4c=2(3)+3\left(-\frac23\right)-4(-1). 2 a + 3 b − 4 c = 2 ( 3 ) + 3 ( − 3 2 ) − 4 ( − 1 ) . = 6 − 2 + 4 = 8. =6-2+4=8. = 6 − 2 + 4 = 8.
Check options
A: 10 10 10
B: 7 7 7
C: 4 4 4
D: 8 8 8
Therefore, the correct answer is Option D .