If \(\frac{d}{dx}f(x) = 4x^3-\frac{3}{x^4}\) such that \(f(2)=0\), then \(f(x)\) is
If \(\frac{d}{dx}f(x) = 4x^3-\frac{3}{x^4}\) such that \(f(2)=0\), then \(f(x)\) is
\(x^4+\frac{1}{x^3}-\frac{129}{8}\)
\(x^3+\frac{1}{x^4}+\frac{129}{8}\)
\(x^4+\frac{1}{x^3}+\frac{129}{8}\)
\(x^3+\frac{1}{x^4}-\frac{129}{8}\)
The Correct Option is A
Solution and Explanation
It is given that,
\(\frac{d}{dx}f(x) = 4x^3-\frac{3}{x^4}\)
∴ Anti derivative of \( 4x^3-\frac{3}{x^4}\) = f(X)
∴ f(x) = \(\int 4x^3-\frac{3}{x^4}dx\)
f(x) = \(4\int x^3dx-3\int (x-4)dx\)
f(x) = \(4\bigg(\frac{x^4}{4}\bigg)-3\frac{x^{-3}}{-3}+C\)
∴ f(x) = x4 + \(\frac{1}{x^3}\) + C
Also,
f(2) = 0
∴ f(2) = \((2)^4+\frac{1}{(2)^3}\)= 0
⇒ 16+\(\frac{1}{8}\)+C = 0
⇒ C = -(16+\(\frac{1}{8}\))
⇒ C = -\(-\frac{129}{8}\)
∴ f(x) = x4 + \(\frac{1}{x^3}-\frac{129}{8}\)
Hence, the correct Answer is A.
Top Questions on integral
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- \( \int_{0}^{\pi/4} \frac{\cos^2 x \sin^2 x}{\left( \cos^3 x + \sin^3 x \right)^2} \, dx \) is equal to:
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- Let \( S = (-1, \infty) \) and \( f : S \rightarrow \mathbb{R} \) be defined as \[ f(x) = \int_{-1}^{x} (e^t - 1)^{11} (2t - 1)^5 (t - 2)^7 (t - 3)^{12} (2t - 10)^{61} \, dt \] Let \( p = \) Sum of squares of the values of \( x \), where \( f(x) \) attains local maxima on \( S \). And \( q = \) Sum of the values of \( x \), where \( f(x) \) attains local minima on \( S \). Then, the value of \( p^2 + 2q \) is ______
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Questions Asked in CBSE CLASS XII exam
- Find the inverse of each of the matrices,if it exists \(\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\)
- For what values of x,\(\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\)\(\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 & 1 \\1&0&2 \end{bmatrix}\)\(\begin{bmatrix} 0 \\2\\x\end{bmatrix}\)=O?
What is the Planning Process?
- CBSE CLASS XII - 2023
- Planning process steps
- Find the inverse of each of the matrices,if it exists. \(\begin{bmatrix} 2 & 3\\ 5 & 7 \end{bmatrix}\)
- Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)
Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.
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