Let $f (x) = \int x^3\sqrt{3-x^2} \, dx$. If $5f (\sqrt{2}) = -4$, then $f (1)$ is equal to
-$\frac{6\sqrt{2}}{5}$
Integral Function Calculation
We need to find the value of $f(1)$ for the integral function $f (x) = \int x^3\sqrt{3-x^2} \, dx$, given the condition $5f (\sqrt{2}) = -4$. This involves integration using substitution and then evaluating the function.
Step 1: Integrate $f(x)$
To solve the integral $f (x) = \int x^3\sqrt{3-x^2} \, dx$, we use the substitution method.
- Let $u = 3-x^2$. Then $du = -2x \, dx$, which means $x \, dx = -\frac{1}{2} du$.
- Also, $x^2 = 3-u$.
- Rewrite the integral in terms of $u$: $f(x) = \int x^2 \sqrt{3-x^2} (x \, dx) = \int (3-u) \sqrt{u} \left(-\frac{1}{2} du\right)$ $f(x) = -\frac{1}{2} \int (3u^{1/2} - u^{3/2}) \, du$
- Integrate with respect to $u$: $f(x) = -\frac{1}{2} \left( 3 \frac{u^{3/2}}{3/2} - \frac{u^{5/2}}{5/2} \right) + C$ $f(x) = -\frac{1}{2} \left( 2 u^{3/2} - \frac{2}{5} u^{5/2} \right) + C$ $f(x) = -u^{3/2} + \frac{1}{5} u^{5/2} + C$
- Substitute back $u = 3-x^2$: $f(x) = -(3-x^2)^{3/2} + \frac{1}{5} (3-x^2)^{5/2} + C$
Step 2: Determine Constant of Integration
Use the given condition $5f (\sqrt{2}) = -4$ to find the value of $C$.
- First, evaluate $f(x)$ at $x = \sqrt{2}$. When $x = \sqrt{2}$, $x^2 = 2$. So, $u = 3 - 2 = 1$.
- Substitute $u=1$ into the expression for $f(x)$: $f(\sqrt{2}) = -(1)^{3/2} + \frac{1}{5} (1)^{5/2} + C = -1 + \frac{1}{5} + C = -\frac{4}{5} + C$
- Apply the condition $5f(\sqrt{2}) = -4$: $5 \left(-\frac{4}{5} + C\right) = -4$ $-4 + 5C = -4$ $5C = 0$ $C = 0$
- Therefore, the specific function is $f(x) = -(3-x^2)^{3/2} + \frac{1}{5} (3-x^2)^{5/2}$.
Step 3: Calculate $f(1)$
Now, evaluate the function $f(x)$ at $x = 1$.
- When $x = 1$, $x^2 = 1$. So, $u = 3 - 1 = 2$.
- Substitute $u=2$ into the expression for $f(x)$: $f(1) = -(2)^{3/2} + \frac{1}{5} (2)^{5/2}$
- Simplify the expression: $f(1) = -2\sqrt{2} + \frac{1}{5} (2^2 \cdot \sqrt{2})$ $f(1) = -2\sqrt{2} + \frac{1}{5} (4\sqrt{2})$ $f(1) = -2\sqrt{2} + \frac{4\sqrt{2}}{5}$
- Combine the terms: $f(1) = \sqrt{2} \left(-2 + \frac{4}{5}\right) = \sqrt{2} \left(\frac{-10 + 4}{5}\right)$ $f(1) = \sqrt{2} \left(-\frac{6}{5}\right)$ $f(1) = -\frac{6\sqrt{2}}{5}$
Similar Questions
Let a line passing through the point $(4, 1, 0)$ intersect the line $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $L_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then \(\begin{bmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{bmatrix}\) is equal to
If $y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix}$, $x \in R$, then $\frac{d^2y}{dx^2} + y$ is equal to
Let $f(x)= \begin{cases} (1+ax)^{1/x} & , x <0 \\ 1+b & , x = 0 \\ \frac{(x+4)^{1/2}-2}{(x+c)^{1/3}-2} & , x > 0 \end{cases}$
be continuous at $x = 0$. Then $e^{abc}$ is equal to:
Let $g$ be a differentiable function such that $\int_0^x g(t)dt=x-\int_0^x tg(t)dt$, $x\ge0$ and let $y = y(x)$ satisfy the differential equation $\frac{dy}{dx} - y\tan x = 2(x+1)\sec x g(x)$, $x\in [0,\frac{\pi}{2})$. If $y(0) = 0$, then $y(\frac{\pi}{3})$ is equal to
The area of the region bounded by the curve $y = \max \{|x|, x|x-2|\}$, the x-axis and the lines $x = -2$ and $x = 4$ is equal to
- If the function $f(x) = 2x^3-9ax^2+12a^2x+1$, where $a > 0$, attains its local maximum and local minimum values at p and q, respectively, such that $p^2 = q$, then $f(3)$ is equal to :
Let $f: R\to R$ be a twice differentiable function such that
$(\sin x \cos y) (f(2x+2y)-f(2x-2y)) = (\cos x \sin y) (f(2x+2y)+f(2x-2y))$, for all $x, y \in R$.
If $f'(0) = \frac{1}{2}$, then the value of $24 f'' (\frac{5\pi}{3})$ is:If the area of the region $\{(x, y):|4-x^2|\le y \le x^2, y\le4,x\ge0)$ is $\left(\frac{80\sqrt{2}}{\alpha}-\beta\right)$, $\alpha, \beta\in N$, then $\alpha+\beta$ is equal to _____________.
- Let $a > 0$. If the function $f(x) = 6x^3-45ax^2+108a^2x+1$ attains its local maximum and minimum values at the points $x_1$ and $x_2$ respectively such that $x_1x_2=54$, then $a + x_1 + x_2$ is equal to:
- Let $f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5$ and $2g(x) - 3g\left(\frac{1}{x}\right) = x, x > 0$. If $\alpha = \int_{1}^{2}f(x) dx$, and $\beta = \int_{1}^{2}g(x) dx$, then the value of $9\alpha + \beta$ is:
Important Questions from Calculus
Let a line passing through the point $(4, 1, 0)$ intersect the line $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $L_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then \(\begin{bmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{bmatrix}\) is equal to
If $y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix}$, $x \in R$, then $\frac{d^2y}{dx^2} + y$ is equal to
Let $f(x)= \begin{cases} (1+ax)^{1/x} & , x <0 \\ 1+b & , x = 0 \\ \frac{(x+4)^{1/2}-2}{(x+c)^{1/3}-2} & , x > 0 \end{cases}$
be continuous at $x = 0$. Then $e^{abc}$ is equal to:
Let $g$ be a differentiable function such that $\int_0^x g(t)dt=x-\int_0^x tg(t)dt$, $x\ge0$ and let $y = y(x)$ satisfy the differential equation $\frac{dy}{dx} - y\tan x = 2(x+1)\sec x g(x)$, $x\in [0,\frac{\pi}{2})$. If $y(0) = 0$, then $y(\frac{\pi}{3})$ is equal to
The area of the region bounded by the curve $y = \max \{|x|, x|x-2|\}$, the x-axis and the lines $x = -2$ and $x = 4$ is equal to
