Find the following integral: \(\int \frac{x^3+3x+4}{\sqrt x}dx\)
Solution and Explanation
\(\int \frac{x^3+3x+4}{\sqrt x}dx\)
= \(\int \bigg(x^{\frac{5}{2}}+3x^{\frac{1}{2}}+4x^{-\frac{1}{2}}\bigg)dx\)
=\(\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+\frac{3\bigg(x^{\frac{3}{2}}\bigg)}{\frac{3}{2}}+\frac{4\bigg(x^{\frac{1}{2}}\bigg)}{\frac{1}{2}}+C\)
=\(\frac{2}{7}x^{\frac{7}{2}}+2x^{\frac{3}{2}}+8x^{\frac{1}{2}}+C\)
= \(\frac{2}{7}x^{\frac{7}{2}}+2x^{\frac{3}{2}}+8\sqrt x+C\)
Top Questions on integral
- Let \( y = f(x) \) be a thrice differentiable function in \( (-5, 5) \). Let the tangents to the curve \( y = f(x) \) at \( (1, f(1)) \) and \( (3, f(3)) \) make angles \( \frac{\pi}{6} \) and \( \frac{\pi}{4} \), respectively, with the positive x-axis. If \(2 \int_{\frac{1}{\sqrt{3}}}^{1} \left( \left( f'(t) \right)^2 + 1 \right) f''(t) \, dt = \alpha + \beta \sqrt{3}\) where \( \alpha \), \( \beta \) are integers, then the value of \( \alpha + \beta \) equals
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- Let \( r_k = \frac{\int_{0}^{1} (1 - x^7)^k \, dx}{\int_{0}^{1} (1 - x^7)^{k+1} \, dx}, \, k \in \mathbb{N} \). Then the value of \[ \sum_{k=1}^{10} \frac{1}{7(r_k - 1)}\]is equal to ______.
Questions Asked in CBSE CLASS XII exam
- Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)
- 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?
- Find the inverse of each of the matrices,if it exists \(\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\)
What is the Planning Process?
- CBSE CLASS XII - 2023
- Planning process steps
- Find the inverse of each of the matrices,if it exists. \(\begin{bmatrix} 2 & 3\\ 5 & 7 \end{bmatrix}\)
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.