KCET · Maths · Definite Integration
\(\int_2^8 \frac{5^{\sqrt{10-x}}}{5^{\sqrt{x}}+5^{\sqrt{10-x}}} d x\) is equals to
- A \(6\)
- B \(4\)
- C \(3\)
- D \(5\)
Answer & Solution
Correct Answer
(C) \(3\)
Step-by-step Solution
Detailed explanation
Let \(I=\int_2^8 \frac{5^{\sqrt{10-x}}}{5^{\sqrt{x}}+5^{\sqrt{10-x}}} d x\)
Using property \(\int_0^a f(x) d x=\int_0^a f(0+a-x) d x\)
\(\begin{aligned} & I=\int_2^8 5^{\sqrt{10-(2+8-x)}} \\ & I=\int_2^8 \frac{5^{\sqrt{10-10+x}}}{5^{\sqrt{10-x}}+5^{\sqrt{10-10+x}}} \\ & I=\int_2^8 \frac{5^{\sqrt{x}}}{5^{\sqrt{10-x}}+5^{\sqrt{x}}}\end{aligned}\)
Adding Eqs. (i) and (ii), we get
\(\begin{aligned} & 2 I=\int_2^8 \frac{5^{\sqrt{10-x}}+5^{\sqrt{x}}}{5^{\sqrt{x}}+5^{\sqrt{10-x}}} d x \\ & I I=\int_2^8 d x \\ & I=\frac{1}{2}[x]_2^8=\frac{1}{2} \times 6=3\end{aligned}\)
Using property \(\int_0^a f(x) d x=\int_0^a f(0+a-x) d x\)
\(\begin{aligned} & I=\int_2^8 5^{\sqrt{10-(2+8-x)}} \\ & I=\int_2^8 \frac{5^{\sqrt{10-10+x}}}{5^{\sqrt{10-x}}+5^{\sqrt{10-10+x}}} \\ & I=\int_2^8 \frac{5^{\sqrt{x}}}{5^{\sqrt{10-x}}+5^{\sqrt{x}}}\end{aligned}\)
Adding Eqs. (i) and (ii), we get
\(\begin{aligned} & 2 I=\int_2^8 \frac{5^{\sqrt{10-x}}+5^{\sqrt{x}}}{5^{\sqrt{x}}+5^{\sqrt{10-x}}} d x \\ & I I=\int_2^8 d x \\ & I=\frac{1}{2}[x]_2^8=\frac{1}{2} \times 6=3\end{aligned}\)
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