Question

Fig.2 shows the curve y = x³ +1. Find the area under the curve shown by the shaded region.

• A. $\frac{5}{4}$ sq units
• B. $\frac{177}{64}$ sq units
• C. $\frac{81}{64}$ sq units
• D. 144 sq units

Answer 1

Answer: (B)

$\frac{177}{64}$ sq units

Answer 2

The shaded region represents the area under the curve $y = x^3 + 1$ from $x = 0$ to $x = 1.5$. To find this area, we need to evaluate the definite integral of the function $y = x^3 + 1$ with respect to $x$ from the lower limit $0$ to the upper limit $1.5$.

The area $A$ is given by:

$A = \int_{0}^{1.5} (x^3 + 1) dx$

First, find the indefinite integral:

$\int (x^3 + 1) dx = \frac{x^{3+1}}{3+1} + x + C = \frac{x^4}{4} + x + C$

Now, evaluate the definite integral using the limits:

$A = \left[ \frac{x^4}{4} + x \right]_{0}^{1.5}$

Substitute the upper limit ($x = 1.5$) and the lower limit ($x = 0$):

$A = \left( \frac{(1.5)^4}{4} + 1.5 \right) - \left( \frac{(0)^4}{4} + 0 \right)$

$A = \frac{(1.5)^4}{4} + 1.5$

Converting $1.5$ to a fraction: $1.5 = \frac{3}{2}$.

$A = \frac{\left(\frac{3}{2}\right)^4}{4} + \frac{3}{2} = \frac{\frac{3^4}{2^4}}{4} + \frac{3}{2} = \frac{\frac{81}{16}}{4} + \frac{3}{2} = \frac{81}{16 \times 4} + \frac{3}{2} = \frac{81}{64} + \frac{3}{2}$

To add the fractions, find a common denominator, which is 64:

$\frac{3}{2} = \frac{3 \times 32}{2 \times 32} = \frac{96}{64}$

$A = \frac{81}{64} + \frac{96}{64} = \frac{81 + 96}{64} = \frac{177}{64}$

Therefore, the area under the curve is $\frac{177}{64}$ square units.