Prove: \(∫^1_{-1}x^{17}cos^{4}xdx\)
Solution and Explanation
Also,let \(ƒ(x)=x^{17}cos^4x\)
\(⇒ƒ(-x)=(-x)^{17}cos^4(-x)=-x^{17}cos^4x=-ƒ(x)\)
Therefore,\(ƒ(x)\)is an odd function.
It is known that if \(f(x)\)is an odd function,then\(∫^a_{-a}ƒ(x)dx=0\)
\(∴I=∫^1_{-1}x^{17}cos^4 xdx=0\)
Hence,the given result is proved.
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Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
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How will the rate of above reaction change if the volume of the reaction vessel is decreased to one third of its original volume? Will there be any change in the order of reaction with the reduced volume?
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(i) What is the probability that selected person is a female?
(ii) If a male person is selected, what is the probability that he will not be suffering from lung problems?
(iii)(a) A person selected at random is detected with lung complications. Find the probability that selected person is a female.
OR
(iii)(b) A person selected at random is not having lung problems. Find the probability that the person is a male.
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Concepts Used:
Definite Integral
Definite integral is an operation on functions which approximates the sum of the values (of the function) weighted by the length (or measure) of the intervals for which the function takes that value.
Definite integrals - Important Formulae Handbook
A real valued function being evaluated (integrated) over the closed interval [a, b] is written as :
\(\int_{a}^{b}f(x)dx\)
Definite integrals have a lot of applications. Its main application is that it is used to find out the area under the curve of a function, as shown below:
