KCET · Maths · Definite Integration
If \(\int_{0}^{1} f(x) d x=5\), then the value of ...+100 \(\int_{0}^{1} \mathrm{x}^{9} \mathrm{f}\left(\mathrm{x}^{10}\right) \mathrm{dx}\) is equal to
- A 125
- B 625
- C 275
- D 55
Answer & Solution
Correct Answer
(D) 55
Step-by-step Solution
Detailed explanation
Given, \(\int_{0}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}=5\)
Let \(I=100 \int_{0}^{1} x^{9} f\left(x^{10}\right) d x\)
Put \(\mathrm{x}^{10}=\mathrm{t} \Rightarrow 10 \mathrm{x}^{9} \mathrm{dx}=\mathrm{dt}\)
\(\therefore \quad \mathrm{I}=100 \int_{0}^{1} \frac{\mathrm{f}(\mathrm{t})}{10} \mathrm{dt}\)
\(=10 \times 5=50\)
\(\therefore \quad \int_{0}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}+100 \int_{0}^{1} \mathrm{x}^{9} \mathrm{f}\left(\mathrm{x}^{10}\right) \mathrm{dx}\)
\(=5+50=55\)
Let \(I=100 \int_{0}^{1} x^{9} f\left(x^{10}\right) d x\)
Put \(\mathrm{x}^{10}=\mathrm{t} \Rightarrow 10 \mathrm{x}^{9} \mathrm{dx}=\mathrm{dt}\)
\(\therefore \quad \mathrm{I}=100 \int_{0}^{1} \frac{\mathrm{f}(\mathrm{t})}{10} \mathrm{dt}\)
\(=10 \times 5=50\)
\(\therefore \quad \int_{0}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}+100 \int_{0}^{1} \mathrm{x}^{9} \mathrm{f}\left(\mathrm{x}^{10}\right) \mathrm{dx}\)
\(=5+50=55\)
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