Qus : 1
Phrases PYQ
2
$\int f(x)\mathrm{d}x=g(x)$, then $\int {x}^5f({x}^3)\mathrm{d}x$
1
$$\frac{1}{3}{x}^3g({x}^3)-3\int {x}^4g({x}^3)\mathrm{d}x+C$$ 2
$$\frac{1}{3}{x}^3g({x}^3)-\int {x}^2g({x}^3)\mathrm{d}x+C$$ 3
$$\frac{1}{3}{x}^3g({x}^3)-\int {x}^3g({x}^3)\mathrm{d}x+C$$ 4
None of these Go to Discussion
Phrases Previous Year PYQ
Phrases NIMCET 2023 PYQ
Solution
Quick Solution
Given:
\( \int f(x)\, dx = g(x) \)
Required: \( \int x^5 f(x^3)\, dx \)
Use substitution:
Let \( u = x^3 \Rightarrow du = 3x^2\, dx \Rightarrow dx = \frac{du}{3x^2} \)
Now rewrite the integral:
\[
\int x^5 f(x^3)\, dx
= \int x^5 f(u) \cdot \frac{du}{3x^2}
= \frac{1}{3} \int x^3 f(u)\, du
\]
But \( x^3 = u \), so:
\[
\frac{1}{3} \int u f(u)\, du
\]
Now integrate by parts or use the identity:
\[
\int u f(u)\, du = u g(u) - \int g(u)\, du
\]
Final answer:
\[
\int x^5 f(x^3)\, dx = \frac{1}{3} \left[ x^3 g(x^3) - \int g(x^3) \cdot 3x^2\, dx \right]
= x^3 g(x^3) - \int x^2 g(x^3)\, dx
\]
\[
\boxed{ \int x^5 f(x^3)\, dx = x^3 g(x^3) - \int x^2 g(x^3)\, dx }
\]
Qus : 4
Phrases PYQ
2
If
, then the values of A
1 , A
2 , A
3 , A
4 are
1
A1
= 1/2 , A2 =1/4 , A3 = 1/6 , A4 = 1/8 2
A1
= 1/8, A2 =1/16 , A3 = 1/24 , A4 = 1/32 3
A1
= 1/6, A2 =1/12 , A3 = 1/18 , A4 = 1/24 4
A1
= 1/4, A2 =1/8 , A3 = 1/12 , A4 = 1/16 Go to Discussion
Phrases Previous Year PYQ
Phrases NIMCET 2019 PYQ
Solution
Qus : 7
Phrases PYQ
3
$\int \log_{10} x , dx$ is
1
$(x - 1)\log e,x + c$ 2
$\log 10 \cdot x \log e\left(\frac{x}{e}\right) + c$ 3
$\log 10 \cdot x \log e\left(\frac{x}{e}\right) + c$ 4
$\dfrac{1}{x} + c$ Go to Discussion
Phrases Previous Year PYQ
Phrases NIMCET 2010 PYQ
Solution Use change of base:
$\log_{10} x = \dfrac{\ln x}{\ln 10}$
$\int \log_{10} x , dx
= \dfrac{1}{\ln 10} \int x(\ln x)' dx
= \dfrac{1}{\ln 10} (x\ln x - x)$
Rewrite:
$= \log 10 \cdot x \log e\left(\frac{x}{e}\right) + c$
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