Given curves:
\( y = x^2 - 5x \quad \text{and} \quad y = 7x - x^2 \)
Let
\( f(x) = x^2 - 5x \quad \text{and} \quad g(x) = 7x - x^2 \)
To find the area enclosed between these curves, we calculate:
\( \int_0^6 (g(x) - f(x)) \, dx = \int_0^6 ((7x - x^2) - (x^2 - 5x)) \, dx \)
Simplify the integrand:
\( = \int_0^6 (12x - 2x^2) \, dx \)
Now, integrate term by term:
\( = \left[ \frac{12x^2}{2} - \frac{2x^3}{3} \right]_0^6 \)
Substitute the limits:
\( = (6 \cdot 6^2) - \frac{2}{3} \cdot (6)^3 \)
\( = 216 - 144 = 72 \, \text{unit}^2 \)
The value of \[ \lim_{x \to \infty} \left( x - \sqrt{x^2 + x} \right) \] is equal to:
Integration of \(\ln(x)\) with \(x\), i.e. \(\int \ln(x)dx =\) __________.
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: