JEE Main Definite Integration Practice Questions With Solutions

If the value of the integral $\underset{-1}{\overset{1}{\int }}\frac{\cos \alpha x}{1+3^x}dx$\int_{- 1}^{1} \frac{cos ⁡ \alpha x}{1 + 3^{x}} d x is $\frac{2}{\pi }$\frac{2}{\pi}.Then, a value of $\alpha$\alpha is

$\text{ Let }f(x)=\{\begin{array}{lr}-2,& -2\le x\le 0\\ x-2,& 0<x\le 2\end{array}\text{ and }\mathrm{h}(x)=f(|x|)+|f(x)|\text{. Then }\underset{-2}{\overset{2}{\int }}\mathrm{h}(x)\mathrm{d}x\text{ is equal to: }$\textrm{ }\text{Let}\textrm{ } f \left(\right. x \left.\right) = \left{\right. - 2 , & - 2 \leq x \leq 0 \\ x - 2 , & 0 < x \leq 2 \textrm{ }\text{and}\textrm{ } h \left(\right. x \left.\right) = f \left(\right. \left|\right. x \left|\right. \left.\right) + \left|\right. f \left(\right. x \left.\right) \left|\right. .\text{ Then}\textrm{ } \int_{- 2}^{2} \textrm{ } h \left(\right. x \left.\right) d x \textrm{ }\text{is equal to}:\textrm{ }

Let $f,g:(0,\mathrm{\infty })\to \mathbb{R}$f , g : \left(\right. 0 , \infty \left.\right) \rightarrow \mathbb{R} be two functions defined by $f(x)=\underset{-x}{\overset{x}{\int }}(|t|-t^2)e^{-t^2}dt$f \left(\right. x \left.\right) = \int_{- x}^{x} \left(\right. \left|\right. t \left|\right. - t^{2} \left.\right) e^{- t^{2}} d t and $g(x)=\underset{0}{\overset{x^2}{\int }}{t}^{1/2}{e}^{-t}dt$g \left(\right. x \left.\right) = \int_{0}^{x^{2}} t^{1 / 2} e^{- t} d t. Then, the value of $9(f\left(\sqrt{{\log }_{e}9}\right)+g\left(\sqrt{{\log }_{e}9}\right))$9 \left(\right. f \left(\right. \sqrt{log_{e} ⁡ 9} \left.\right) + g \left(\right. \sqrt{log_{e} ⁡ 9} \left.\right) \left.\right) is equal to :

Let $f:\mathbb{R}\to \mathbb{R}$f : \mathbb{R} \rightarrow \mathbb{R} be a function defined by $f(x)=\frac{x}{{(1+x^4)}^{1/4}}$f \left(\right. x \left.\right) = \frac{x}{\left(\left(\right. 1 + x^{4} \left.\right)\right)^{1 / 4}}, and $g(x)=f(f(f(f(x))))$g \left(\right. x \left.\right) = f \left(\right. f \left(\right. f \left(\right. f \left(\right. x \left.\right) \left.\right) \left.\right) \left.\right). Then, $18{\int }_0^{\sqrt{2\sqrt{5}}}{x}^{2}g(x)dx$18 \int_{0}^{\sqrt{2 \sqrt{5}}} x^{2} g \left(\right. x \left.\right) d x is equal to

Let $y=f(x)$y = f \left(\right. x \left.\right) be a thrice differentiable function in $(-5,5)$\left(\right. - 5 , 5 \left.\right). Let the tangents to the curve $y=f(x)$y = f \left(\right. x \left.\right) at $(1,f(1))$\left(\right. 1 , f \left(\right. 1 \left.\right) \left.\right) and $(3,f(3))$\left(\right. 3 , f \left(\right. 3 \left.\right) \left.\right) make angles $\pi /6$\pi / 6 and $\pi /4$\pi / 4, respectively with positive $x$x-axis. If $27\underset{1}{\overset{3}{\int }}({(f^{\prime }(t))}^2+1)f^{\prime \prime }(t)dt=\alpha +\beta \sqrt{3}$27 \int_{1}^{3} \left(\right. \left(\left(\right. f^{′} \left(\right. t \left.\right) \left.\right)\right)^{2} + 1 \left.\right) f^{′ ′} \left(\right. t \left.\right) d t = \alpha + \beta \sqrt{3} where $\alpha ,\beta$\alpha , \beta are integers, then the value of $\alpha +\beta$\alpha + \beta equals

Let $a$a and $b$b be real constants such that the function $f$f defined by $f(x)=\{\begin{array}{ll}{x}^2+3x+a& ,x\le 1\\ bx+2& ,x>1\end{array}$f \left(\right. x \left.\right) = \left{\right. x^{2} + 3 x + a & , x \leq 1 \\ b x + 2 & , x > 1 be differentiable on $\mathbb{R}$\mathbb{R}. Then, the value of $\underset{-2}{\overset{2}{\int }}f(x)dx$\int_{- 2}^{2} f \left(\right. x \left.\right) d x equals

Let $\mathrm{f}:\mathbb{R}\to \mathbb{R}$f : \mathbb{R} \rightarrow \mathbb{R} be defined as $f(x)=a{e}^{2x}+b{e}^x+cx$f \left(\right. x \left.\right) = a e^{2 x} + b e^{x} + c x. If $f(0)=-1,f^{\prime }({\log }_{e}2)=21$f \left(\right. 0 \left.\right) = - 1 , f^{′} \left(\right. log_{e} ⁡ 2 \left.\right) = 21 and ${\int }_0^{{\log }_{e}4}(f(x)-cx)dx=\frac{39}{2}$\int_{0}^{log_{e} ⁡ 4} \left(\right. f \left(\right. x \left.\right) - c x \left.\right) d x = \frac{39}{2}, then the value of $|a+b+c|$\left|\right. a + b + c \left|\right. equals