Question 12 Marks
Read the following text and answer the following questions on the basis of the same
Mohan is the student of class XII of Sunshine School. In the mathematics class, teacher defines the method for evaluation of definite integral, she said, To evaluate the definite integral $\underset{\pi}{\equiv} f(x) d x$ a continuous function f(x) defined on [a, b], we may use the following algorithm.
Step 1: Find the indefinite integral $\stackrel{b}{\equiv} f(x) d x$
Let this be P(x) There is no need to keep the constant of integration.
Step 2: Evaluate $\phi(b)$ and $\phi(a)$.
Step 3: Calculate $\phi(b)-\phi(a)$.
The number obtained in step 3 is the value of definite integral $\int_a^b f(x) d x$
Q.1. Evaluate: $\int_0^1 \frac{1}{\sqrt{1+x}+\sqrt{x}} d x$
Q.2. If $\int_1^a\left(3 x^2+2 x+1\right) d x=1$, find real values of $a$.
Mohan is the student of class XII of Sunshine School. In the mathematics class, teacher defines the method for evaluation of definite integral, she said, To evaluate the definite integral $\underset{\pi}{\equiv} f(x) d x$ a continuous function f(x) defined on [a, b], we may use the following algorithm.
Step 1: Find the indefinite integral $\stackrel{b}{\equiv} f(x) d x$
Let this be P(x) There is no need to keep the constant of integration.
Step 2: Evaluate $\phi(b)$ and $\phi(a)$.
Step 3: Calculate $\phi(b)-\phi(a)$.
The number obtained in step 3 is the value of definite integral $\int_a^b f(x) d x$
Q.1. Evaluate: $\int_0^1 \frac{1}{\sqrt{1+x}+\sqrt{x}} d x$
Q.2. If $\int_1^a\left(3 x^2+2 x+1\right) d x=1$, find real values of $a$.
Answer
View full question & answer→(1)
$\int_0^1 \frac{1}{\sqrt{1+x}+\sqrt{x}} d x$
$\begin{array}{l}=\int_0^1 \frac{\sqrt{1+x}-\sqrt{x}}{(\sqrt{1+x}+\sqrt{x})(\sqrt{1+x}-\sqrt{x})} \\ =\int_0^1(\sqrt{1+x}-\sqrt{x}) d x \\ =\left[\frac{2}{3}(1+x)^{\frac{3}{2}}-\frac{2}{3} x^{\frac{3}{2}}\right]_0^1\end{array}$
$\begin{array}{l}=\left[\frac{2}{3}(1+1)^{\frac{3}{2}}-\frac{2}{3}(1)^{\frac{3}{2}}\right]-\left[\frac{2}{3}(1+0)^{\frac{3}{2}}-\frac{2}{3}(0)^{\frac{3}{2}}\right] \\ =\frac{2}{3}\left(2^{\frac{3}{2}}-1\right)-\frac{2}{3}(1-0) \\ =\frac{2}{3}(2 \sqrt{2}-2)=\frac{4}{3}(\sqrt{2}-1)\end{array}$
(2)
Given $\int_1^a\left(3 x^2+2 x+1\right) d x=11$
$\Rightarrow \quad\left[x^3+x^2+x\right]_1^a=11$
$\Rightarrow \quad\left(a^3+a^2+a\right)-(1+1+1)=11$
$\Rightarrow \quad a^3+a^2+a-3=11$
$\Rightarrow \quad a^3+a^2+a-14=0$
$\Rightarrow \quad(a-2)\left(a^2+3 a+7\right)=0$
$\Rightarrow \quad a=2$
$\left[\because a^2+3 a+7 \neq 0\right.$, for any $\left.a \in R\right]$
$\int_0^1 \frac{1}{\sqrt{1+x}+\sqrt{x}} d x$
$\begin{array}{l}=\int_0^1 \frac{\sqrt{1+x}-\sqrt{x}}{(\sqrt{1+x}+\sqrt{x})(\sqrt{1+x}-\sqrt{x})} \\ =\int_0^1(\sqrt{1+x}-\sqrt{x}) d x \\ =\left[\frac{2}{3}(1+x)^{\frac{3}{2}}-\frac{2}{3} x^{\frac{3}{2}}\right]_0^1\end{array}$
$\begin{array}{l}=\left[\frac{2}{3}(1+1)^{\frac{3}{2}}-\frac{2}{3}(1)^{\frac{3}{2}}\right]-\left[\frac{2}{3}(1+0)^{\frac{3}{2}}-\frac{2}{3}(0)^{\frac{3}{2}}\right] \\ =\frac{2}{3}\left(2^{\frac{3}{2}}-1\right)-\frac{2}{3}(1-0) \\ =\frac{2}{3}(2 \sqrt{2}-2)=\frac{4}{3}(\sqrt{2}-1)\end{array}$
(2)
Given $\int_1^a\left(3 x^2+2 x+1\right) d x=11$
$\Rightarrow \quad\left[x^3+x^2+x\right]_1^a=11$
$\Rightarrow \quad\left(a^3+a^2+a\right)-(1+1+1)=11$
$\Rightarrow \quad a^3+a^2+a-3=11$
$\Rightarrow \quad a^3+a^2+a-14=0$
$\Rightarrow \quad(a-2)\left(a^2+3 a+7\right)=0$
$\Rightarrow \quad a=2$
$\left[\because a^2+3 a+7 \neq 0\right.$, for any $\left.a \in R\right]$
