If for every integer n, ∫nn + 1fxdx = n2, then the value of ∫- 24f(x)dx is
16
14
19
None of these
The value of integral ∫0πxfsinxdx is
0
π∫0π2fsinxdx
π4∫0πfsinxdx
Solution of Given equation is dydx = xlogx2 + xsiny + ycosy⇒ siny + ycosydy = xlogx2 + xdxOn integrating both sides, we get ∫siny + ycosydy = ∫xlogx2 + xdx⇒ - cosy + ysiny + cosy = x22logx2 is
ysiny = x2logx + C
ysiny = x2 + C
ysiny = x2 + logx + C
ysiny = xogx + C
limx→∞∫02xxexdxe4x2 equals
∞
2
12
If xpyq = (x + y)p + q, then dydx is equal to
yx
pyqx
xy
qypx
A.
Given, xpyq = (x + y)p + q
Taking log on both sides, we get
p logx + q logy = p + q logx + y
On differentiating w.r.t.x, we get
px + qydydx = p + qx + y1 + dydx⇒ dydxqy - p + qx + y = p + qx + y - px⇒ dydxqx + qy - px - pyyx + y = px + qx - px - pyxx + y⇒ dydx = yx + yqx - py . qx - pyxx + y⇒ dydx = yx
Area bounded by the curve y2 = 16x and line y = mx is 23, then m is equal to
3
4
1
The equation zz + az + az + b = 0, represents a circle, if
a2 = b
a2 > b
a2 < b
None of the above
Area included between curves y = x2 - 3x + 2 and y = - x2 + 3x - 2 is
16 sq unit
12 sq unit
1 sq unit
13 sq unit
The solution of dydx + ytanx = secx is
ysecx = tanx + C
ytanx = secx + C
tanx = ytanx + C
xsecx = ytany + C
If y = tan-1sinx + cosxcosx - sinx, then dydx is
π4